\displaystyle \textbf{Question 1: }~\text{Evaluate the following integrals (1--44):}

\displaystyle \text{(1): }\int \left(3x\sqrt{x}+4\sqrt{x}+5\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(3x^{3/2}+4x^{1/2}+5\right)\,dx
\displaystyle = \int 3x^{3/2}\,dx+\int 4x^{1/2}\,dx+\int 5\,dx
\displaystyle = 3\cdot\frac{x^{3/2+1}}{3/2+1}+4\cdot\frac{x^{1/2+1}}{1/2+1}+5x+C
\displaystyle = 3\cdot\frac{x^{5/2}}{5/2}+4\cdot\frac{x^{3/2}}{3/2}+5x+C
\displaystyle = \frac{6}{5}x^{5/2}+\frac{8}{3}x^{3/2}+5x+C
\\

\displaystyle \text{(2): }\int \left(2^x+\frac{5}{x}-\frac{1}{x^{1/3}}\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int 2^x\,dx+\int \frac{5}{x}\,dx-\int x^{-1/3}\,dx
\displaystyle = \frac{2^x}{\log 2}+5\log x-\frac{x^{-1/3+1}}{-1/3+1}+C
\displaystyle = \frac{2^x}{\log 2}+5\log x-\frac{x^{2/3}}{2/3}+C
\displaystyle = \frac{2^x}{\log 2}+5\log x-\frac{3}{2}x^{2/3}+C
\\

\displaystyle \text{(3): }\int \left\{\sqrt{x}\left(ax^2+bx+c\right)\right\}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int x^{1/2}\left(ax^2+bx+c\right)\,dx
\displaystyle = \int \left(ax^{5/2}+bx^{3/2}+cx^{1/2}\right)\,dx
\displaystyle = a\int x^{5/2}\,dx+b\int x^{3/2}\,dx+c\int x^{1/2}\,dx
\displaystyle = a\cdot\frac{x^{5/2+1}}{5/2+1}+b\cdot\frac{x^{3/2+1}}{3/2+1}+c\cdot\frac{x^{1/2+1}}{1/2+1}+C
\displaystyle = a\cdot\frac{x^{7/2}}{7/2}+b\cdot\frac{x^{5/2}}{5/2}+c\cdot\frac{x^{3/2}}{3/2}+C
\displaystyle = \frac{2a}{7}x^{7/2}+\frac{2b}{5}x^{5/2}+\frac{2c}{3}x^{3/2}+C
\\

\displaystyle \text{(4): }\int (2-3x)(3+2x)(1-2x)\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(12x^3+4x^2-17x+6\right)\,dx
\displaystyle = 12\int x^3\,dx+4\int x^2\,dx-17\int x\,dx+\int 6\,dx
\displaystyle = 12\cdot\frac{x^{3+1}}{3+1}+4\cdot\frac{x^{2+1}}{2+1}-17\cdot\frac{x^{1+1}}{1+1}+6x+C
\displaystyle = 3x^4+\frac{4}{3}x^3-\frac{17}{2}x^2+6x+C
\\

\displaystyle \text{(5): }\int \left(\frac{m}{x}+\frac{x}{m}+m^x+x^m+mx\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int\left(\frac{m}{x}+\frac{x}{m}+m^x+x^m+mx\right)\,dx = m\int\frac{1}{x}\,dx+\frac{1}{m}\int x\,dx+\int m^x\,dx+\int x^m\,dx+m\int x\,dx
\displaystyle = m\log|x|+\frac{1}{m}\left(\frac{x^2}{2}\right)+\frac{m^x}{\ln m}+\frac{x^{m+1}}{m+1}+m\left(\frac{x^2}{2}\right)+C
\displaystyle = m\log|x|+\frac{x^2}{2m}+\frac{m^x}{\ln m}+\frac{x^{m+1}}{m+1}+\frac{mx^2}{2}+C
\\

\displaystyle \text{(6): }\int \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\,dx.
\displaystyle \text{Answer:}
\displaystyle \int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 dx = \int\left(x+\frac{1}{x}-2\right) dx
\displaystyle = \int x\,dx+\int\frac{dx}{x}-2\int dx
\displaystyle = \frac{x^2}{2}+\ln|x|-2x+C
\\

\displaystyle \text{(7): }\int \frac{(1+x)^3}{\sqrt{x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int (1+x)^3x^{-1/2}\,dx
\displaystyle = \int \left(1+3x+3x^2+x^3\right)x^{-1/2}\,dx
\displaystyle = \int \left(x^{-1/2}+3x^{1/2}+3x^{3/2}+x^{5/2}\right)\,dx
\displaystyle = \frac{x^{1/2}}{1/2}+3\cdot\frac{x^{3/2}}{3/2}+3\cdot\frac{x^{5/2}}{5/2}+\frac{x^{7/2}}{7/2}+C
\displaystyle = 2x^{1/2}+2x^{3/2}+\frac{6}{5}x^{5/2}+\frac{2}{7}x^{7/2}+C
\\

\displaystyle \text{(8): }\int \left\{x^2+e^{\log x}+\left(\frac{e}{2}\right)^x\right\}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(x^2+x+\left(\frac{e}{2}\right)^x\right)\,dx \quad (\because~e^{\log x}=x)
\displaystyle = \int x^2\,dx+\int x\,dx+\int \left(\frac{e}{2}\right)^x\,dx
\displaystyle = \frac{x^{2+1}}{2+1}+\frac{x^{1+1}}{1+1}+\frac{\left(\frac{e}{2}\right)^x}{\log\left(\frac{e}{2}\right)}+C
\displaystyle = \frac{x^3}{3}+\frac{x^2}{2}+\frac{\left(\frac{e}{2}\right)^x}{\log\left(\frac{e}{2}\right)}+C
\\

\displaystyle \text{(9): }\int \left(x^e+e^x+e^e\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int x^e\,dx+\int e^x\,dx+\int e^e\,dx
\displaystyle = \frac{x^{e+1}}{e+1}+e^x+e^e x+C
\\

\displaystyle \text{(10): }\int \sqrt{x}\left(x^3-\frac{2}{x}\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int x^{1/2}\left(x^3-\frac{2}{x}\right)\,dx
\displaystyle = \int \left(x^{7/2}-2x^{-1/2}\right)\,dx
\displaystyle = \frac{x^{7/2+1}}{7/2+1}-2\cdot\frac{x^{-1/2+1}}{-1/2+1}+C
\displaystyle = \frac{x^{9/2}}{9/2}-2\cdot\frac{x^{1/2}}{1/2}+C
\displaystyle = \frac{2}{9}x^{9/2}-4\sqrt{x}+C
\\

\displaystyle \text{(11): }\int \frac{1}{\sqrt{x}}\left(1+\frac{1}{x}\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(x^{-1/2}+x^{-3/2}\right)\,dx
\displaystyle = \frac{x^{-1/2+1}}{-1/2+1}+\frac{x^{-3/2+1}}{-3/2+1}+C
\displaystyle = \frac{x^{1/2}}{1/2}+\frac{x^{-1/2}}{-1/2}+C
\displaystyle = 2\sqrt{x}-\frac{2}{\sqrt{x}}+C
\\

\displaystyle \text{(12): }\int \frac{x^6+1}{x^2+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(x^4-x^2+1\right)\,dx \quad \left(\because~\frac{x^6+1}{x^2+1}=x^4-x^2+1\right)
\displaystyle = \int x^4\,dx-\int x^2\,dx+\int 1\,dx
\displaystyle = \frac{x^{4+1}}{4+1}-\frac{x^{2+1}}{2+1}+x+C
\displaystyle = \frac{x^5}{5}-\frac{x^3}{3}+x+C
\\

\displaystyle \text{(13): }\int \frac{x^{-1/3}+\sqrt{x}+2}{\sqrt[3]{x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{x^{-1/3}+x^{1/2}+2}{x^{1/3}}\,dx
\displaystyle = \int \left(x^{-2/3}+x^{1/6}+2x^{-1/3}\right)\,dx
\displaystyle = \frac{x^{-2/3+1}}{-2/3+1}+\frac{x^{1/6+1}}{1/6+1}+2\cdot\frac{x^{-1/3+1}}{-1/3+1}+C
\displaystyle = \frac{x^{1/3}}{1/3}+\frac{x^{7/6}}{7/6}+2\cdot\frac{x^{2/3}}{2/3}+C
\displaystyle = 3x^{1/3}+\frac{6}{7}x^{7/6}+3x^{2/3}+C
\\

\displaystyle \text{(14): }\int \frac{(1+\sqrt{x})^2}{\sqrt{x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{1+2\sqrt{x}+x}{\sqrt{x}}\,dx
\displaystyle = \int \left(x^{-1/2}+2+x^{1/2}\right)\,dx
\displaystyle = \frac{x^{-1/2+1}}{-1/2+1}+2x+\frac{x^{1/2+1}}{1/2+1}+C
\displaystyle = \frac{x^{1/2}}{1/2}+2x+\frac{x^{3/2}}{3/2}+C
\displaystyle = 2\sqrt{x}+2x+\frac{2}{3}x^{3/2}+C
\\

\displaystyle \text{(15): }\int \sqrt{x}(3-5x)\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int x^{1/2}(3-5x)\,dx
\displaystyle = \int \left(3x^{1/2}-5x^{3/2}\right)\,dx
\displaystyle = 3\cdot\frac{x^{1/2+1}}{1/2+1}-5\cdot\frac{x^{3/2+1}}{3/2+1}+C
\displaystyle = 3\cdot\frac{x^{3/2}}{3/2}-5\cdot\frac{x^{5/2}}{5/2}+C
\displaystyle = 2x^{3/2}-2x^{5/2}+C
\\

\displaystyle \text{(16): }\int \frac{(x+1)(x-2)}{\sqrt{x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{x^2-x-2}{\sqrt{x}}\,dx
\displaystyle = \int \left(x^{3/2}-x^{1/2}-2x^{-1/2}\right)\,dx
\displaystyle = \frac{x^{3/2+1}}{3/2+1}-\frac{x^{1/2+1}}{1/2+1}-2\cdot\frac{x^{-1/2+1}}{-1/2+1}+C
\displaystyle = \frac{x^{5/2}}{5/2}-\frac{x^{3/2}}{3/2}-2\cdot\frac{x^{1/2}}{1/2}+C
\displaystyle = \frac{2}{5}x^{5/2}-\frac{2}{3}x^{3/2}-4\sqrt{x}+C
\\

\displaystyle \text{(17): }\int \frac{x^5+x^{-2}+2}{x^2}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(x^3+x^{-4}+2x^{-2}\right)\,dx
\displaystyle = \frac{x^{3+1}}{3+1}+\frac{x^{-4+1}}{-4+1}+2\cdot\frac{x^{-2+1}}{-2+1}+C
\displaystyle = \frac{x^4}{4}-\frac{x^{-3}}{3}-2x^{-1}+C
\displaystyle = \frac{x^4}{4}-\frac{1}{3x^3}-\frac{2}{x}+C
\\

\displaystyle \text{(18): }\int (3x+4)^2\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(9x^2+24x+16\right)\,dx
\displaystyle = 9\int x^2\,dx+24\int x\,dx+\int 16\,dx
\displaystyle = 9\cdot\frac{x^{2+1}}{2+1}+24\cdot\frac{x^{1+1}}{1+1}+16x+C
\displaystyle = 3x^3+12x^2+16x+C
\\

\displaystyle \text{(19): }\int \frac{2x^4+7x^3+6x^2}{x^2+2x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{x^2(2x^2+7x+6)}{x(x+2)}\,dx
\displaystyle = \int \frac{x(2x^2+7x+6)}{x+2}\,dx
\displaystyle = \int \frac{2x^3+7x^2+6x}{x+2}\,dx
\displaystyle = \int (2x^2+3x)\,dx
\displaystyle = 2\cdot\frac{x^{2+1}}{2+1}+3\cdot\frac{x^{1+1}}{1+1}+C
\displaystyle = \frac{2}{3}x^3+\frac{3}{2}x^2+C
\\

\displaystyle \text{(20): }\int \frac{5x^4+12x^3+7x^2}{x^2+x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{x^2(5x^2+12x+7)}{x(x+1)}\,dx
\displaystyle = \int \frac{x(5x^2+12x+7)}{x+1}\,dx
\displaystyle = \int \frac{5x^3+12x^2+7x}{x+1}\,dx
\displaystyle = \int (5x^2+7x)\,dx
\displaystyle = 5\cdot\frac{x^{2+1}}{2+1}+7\cdot\frac{x^{1+1}}{1+1}+C
\displaystyle = \frac{5}{3}x^3+\frac{7}{2}x^2+C
\\

\displaystyle \text{(21): }\int \frac{\sin^2 x}{1+\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{1-\cos^2 x}{1+\cos x}\,dx
\displaystyle = \int (1-\cos x)\,dx
\displaystyle = \int 1\,dx-\int \cos x\,dx
\displaystyle = x-\sin x+C
\\

\displaystyle \text{(22): }\int \left(\sec^2 x+\mathrm{cosec}^2 x\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \sec^2 x\,dx+\int \mathrm{cosec}^2 x\,dx
\displaystyle = \tan x-\cot x+C
\\

\displaystyle \text{(23): }\int \frac{\sin^3 x-\cos^3 x}{\sin^2 x\cos^2 x}\,dx. 
\displaystyle \text{Answer:}
\displaystyle = \int \left(\frac{\sin^3 x}{\sin^2 x\cos^2 x}-\frac{\cos^3 x}{\sin^2 x\cos^2 x}\right)\,dx
\displaystyle = \int \left(\frac{\sin x}{\cos^2 x}-\frac{\cos x}{\sin^2 x}\right)\,dx
\displaystyle = \int \left(\sec x\tan x-\mathrm{cosec}\,x\cot x\right)\,dx
\displaystyle = \sec x+\mathrm{cosec}\,x+C
\\

\displaystyle \text{(24): }\int \frac{5\cos^3 x+6\sin^3 x}{2\sin^2 x\cos^2 x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \frac{1}{2}\int \left(\frac{5\cos^3 x}{\sin^2 x\cos^2 x}+\frac{6\sin^3 x}{\sin^2 x\cos^2 x}\right)\,dx
\displaystyle = \frac{1}{2}\int \left(\frac{5\cos x}{\sin^2 x}+\frac{6\sin x}{\cos^2 x}\right)\,dx
\displaystyle = \int \left(\frac{5}{2}\,\mathrm{cosec}\,x\cot x+3\,\sec x\tan x\right)\,dx
\displaystyle = -\frac{5}{2}\,\mathrm{cosec}\,x+3\sec x+C
\\

\displaystyle \text{(25): }\int (\tan x+\cot x)^2\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(\tan^2 x+\cot^2 x+2\tan x\cot x\right)\,dx
\displaystyle = \int \left(\tan^2 x+\cot^2 x+2\right)\,dx \quad (\because~\tan x\cot x=1)
\displaystyle = \int \left((\sec^2 x-1)+(\mathrm{cosec}^2 x-1)+2\right)\,dx
\displaystyle = \int \left(\sec^2 x+\mathrm{cosec}^2 x\right)\,dx
\displaystyle = \tan x-\cot x+C
\\

\displaystyle \text{(26): }\int \frac{1-\cos 2x}{1+\cos 2x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{2\sin^2 x}{2\cos^2 x}\,dx \quad (\because~1-\cos 2x=2\sin^2 x,\;1+\cos 2x=2\cos^2 x)
\displaystyle = \int \tan^2 x\,dx
\displaystyle = \int (\sec^2 x-1)\,dx
\displaystyle = \tan x-x+C
\\

\displaystyle \text{(27): }\int \frac{\cos x}{1-\cos x}\,dx\;\text{ or }\;\int \frac{\cot x}{\mathrm{cosec}\,x-\cot x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(\frac{1}{1-\cos x}-1\right)\,dx
\displaystyle = \int \frac{1}{1-\cos x}\,dx-\int 1\,dx
\displaystyle = \int \frac{1+\cos x}{1-\cos^2 x}\,dx-x \quad \left(\because~\frac{1}{1-\cos x}=\frac{1+\cos x}{(1-\cos x)(1+\cos x)}\right)
\displaystyle = \int \left(\frac{1}{\sin^2 x}+\frac{\cos x}{\sin^2 x}\right)\,dx-x
\displaystyle = \int \left(\mathrm{cosec}^2 x+\mathrm{cosec}\,x\cot x\right)\,dx-x
\displaystyle = -\cot x-\mathrm{cosec}\,x-x+C
\\

\displaystyle \text{(28): }\int \frac{\cos^2 x-\sin^2 x}{\sqrt{1+\cos 4x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{\cos 2x}{\sqrt{1+\cos 4x}}\,dx \quad (\because~\cos^2 x-\sin^2 x=\cos 2x)
\displaystyle = \int \frac{\cos 2x}{\sqrt{2\cos^2 2x}}\,dx \quad (\because~1+\cos 4x=2\cos^2 2x)
\displaystyle = \int \frac{\cos 2x}{\sqrt{2}\cos 2x}\,dx
\displaystyle = \int \frac{1}{\sqrt{2}}\,dx
\displaystyle = \frac{x}{\sqrt{2}}+C
\\

\displaystyle \text{(29): }\int \frac{1}{1-\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{1+\cos x}{1-\cos^2 x}\,dx
\displaystyle = \int \frac{1+\cos x}{\sin^2 x}\,dx
\displaystyle = \int \left(\mathrm{cosec}^2 x+\mathrm{cosec}\,x\cot x\right)\,dx
\displaystyle = -\cot x-\mathrm{cosec}\,x+C
\\

\displaystyle \text{(30): }\int \frac{1}{1-\sin x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{1+\sin x}{1-\sin^2 x}\,dx
\displaystyle = \int \frac{1+\sin x}{\cos^2 x}\,dx
\displaystyle = \int \left(\sec^2 x+\frac{\sin x}{\cos^2 x}\right)\,dx
\displaystyle = \int \left(\sec^2 x+\sec x\tan x\right)\,dx
\displaystyle = \tan x+\sec x+C
\\

\displaystyle \text{(31): }\int \frac{\tan x}{\sec x+\tan x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{\tan x(\sec x-\tan x)}{(\sec x+\tan x)(\sec x-\tan x)}\,dx
\displaystyle = \int \tan x(\sec x-\tan x)\,dx \quad (\because~(\sec x+\tan x)(\sec x-\tan x)=1)
\displaystyle = \int (\sec x\tan x-\tan^2 x)\,dx
\displaystyle = \int \sec x\tan x\,dx-\int (\sec^2 x-1)\,dx
\displaystyle = \sec x-(\tan x-x)+C
\displaystyle = \sec x-\tan x+x+C
\\

\displaystyle \text{(32): }\int \frac{\mathrm{cosec}\,x}{\mathrm{cosec}\,x-\cot x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{\mathrm{cosec}\,x(\mathrm{cosec}\,x+\cot x)}{(\mathrm{cosec}\,x-\cot x)(\mathrm{cosec}\,x+\cot x)}\,dx
\displaystyle = \int \mathrm{cosec}\,x(\mathrm{cosec}\,x+\cot x)\,dx \quad (\because~(\mathrm{cosec}^2 x-\cot^2 x)=1)
\displaystyle = \int \left(\mathrm{cosec}^2 x+\mathrm{cosec}\,x\cot x\right)\,dx
\displaystyle = -\cot x-\mathrm{cosec}\,x+C
\\

\displaystyle \text{(33): }\int \frac{1}{1+\cos 2x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{1}{2\cos^2 x}\,dx \quad (\because~1+\cos 2x=2\cos^2 x)
\displaystyle = \frac{1}{2}\int \sec^2 x\,dx
\displaystyle = \frac{1}{2}\tan x+C
\\

\displaystyle \text{(34): }\int \frac{1}{1-\cos 2x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{1}{2\sin^2 x}\,dx \quad (\because~1-\cos 2x=2\sin^2 x)
\displaystyle = \frac{1}{2}\int \mathrm{cosec}^2 x\,dx
\displaystyle = -\frac{1}{2}\cot x+C
\\

\displaystyle \text{(35): }\int \tan^{-1}\!\left(\frac{\sin 2x}{1+\cos 2x}\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \tan^{-1}(\tan x)\,dx \quad \left(\because~\frac{\sin 2x}{1+\cos 2x}=\tan x\right)
\displaystyle = \int x\,dx
\displaystyle = \frac{x^2}{2}+C
\\

\displaystyle \text{(36): }\int \cos^{-1}(\sin x)\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \cos^{-1}\!\left(\cos\left(\frac{\pi}{2}-x\right)\right)\,dx \quad (\because~\sin x=\cos(\frac{\pi}{2}-x))
\displaystyle = \int \left(\frac{\pi}{2}-x\right)\,dx
\displaystyle = \frac{\pi x}{2}-\frac{x^2}{2}+C
\\

\displaystyle \text{(37): }\int \cot^{-1}\!\left(\frac{\sin 2x}{1-\cos 2x}\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \cot^{-1}(\cot x)\,dx \quad \left(\because~\frac{\sin 2x}{1-\cos 2x}=\cot x\right)
\displaystyle = \int x\,dx
\displaystyle = \frac{x^2}{2}+C
\\

\displaystyle \text{(38): }\int \sin^{-1}\!\left(\frac{2\tan x}{1+\tan^2 x}\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \sin^{-1}(\sin 2x)\,dx \quad \left(\because~\frac{2\tan x}{1+\tan^2 x}=\sin 2x\right)
\displaystyle = \int 2x\,dx
\displaystyle = x^2+C
\\

\displaystyle \text{(39): }\int \frac{(x^3+8)(x-1)}{x^2-2x+4}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{(x+2)(x^2-2x+4)(x-1)}{x^2-2x+4}\,dx \quad (\because~x^3+8=(x+2)(x^2-2x+4))
\displaystyle = \int (x+2)(x-1)\,dx
\displaystyle = \int (x^2+x-2)\,dx
\displaystyle = \frac{x^{2+1}}{2+1}+\frac{x^{1+1}}{1+1}-2x+C
\displaystyle = \frac{x^3}{3}+\frac{x^2}{2}-2x+C
\\

\displaystyle \text{(40): }\int (a\tan x+b\cot x)^2\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(a^2\tan^2 x+b^2\cot^2 x+2ab\tan x\cot x\right)\,dx
\displaystyle = \int \left(a^2\tan^2 x+b^2\cot^2 x+2ab\right)\,dx \quad (\because~\tan x\cot x=1)
\displaystyle = \int \left(a^2(\sec^2 x-1)+b^2(\mathrm{cosec}^2 x-1)+2ab\right)\,dx
\displaystyle = a^2\int \sec^2 x\,dx-a^2\int 1\,dx+b^2\int \mathrm{cosec}^2 x\,dx-b^2\int 1\,dx+2ab\int 1\,dx
\displaystyle = a^2\tan x-a^2x-b^2\cot x-b^2x+2abx+C
\displaystyle = a^2\tan x-b^2\cot x-(a-b)^2x+C
\\

\displaystyle \text{(41): }\int \frac{x^3-3x^2+5x-7+x^2a^x}{2x^2}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \frac{1}{2}\int \left(\frac{x^3}{x^2}-\frac{3x^2}{x^2}+\frac{5x}{x^2}-\frac{7}{x^2}+\frac{x^2a^x}{x^2}\right)\,dx
\displaystyle = \frac{1}{2}\int \left(x-3+\frac{5}{x}-\frac{7}{x^2}+a^x\right)\,dx
\displaystyle = \frac{1}{2}\left(\int x\,dx-3\int 1\,dx+5\int \frac{1}{x}\,dx-7\int x^{-2}\,dx+\int a^x\,dx\right)
\displaystyle = \frac{1}{2}\left(\frac{x^2}{2}-3x+5\log x-7\cdot\frac{x^{-1}}{-1}+\frac{a^x}{\log a}\right)+C
\displaystyle = \frac{1}{2}\left(\frac{x^2}{2}-3x+5\log x+\frac{7}{x}+\frac{a^x}{\log a}\right)+C
\displaystyle = \frac{x^2}{4}-\frac{3x}{2}+\frac{5}{2}\log x+\frac{7}{2x}+\frac{a^x}{2\log a}+C
\\

\displaystyle \text{(42): }\int \frac{\cos x}{1+\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(1-\frac{1}{1+\cos x}\right)\,dx
\displaystyle = x-\int \frac{1}{1+\cos x}\,dx
\displaystyle = x-\int \frac{1-\cos x}{1-\cos^2 x}\,dx
\displaystyle = x-\int \left(\frac{1}{\sin^2 x}-\frac{\cos x}{\sin^2 x}\right)\,dx
\displaystyle = x-\int \left(\mathrm{cosec}^2 x-\mathrm{cosec}\,x\cot x\right)\,dx
\displaystyle = x-\left(-\cot x+\mathrm{cosec}\,x\right)+C
\displaystyle = x+\cot x-\mathrm{cosec}\,x+C
\\

\displaystyle \text{(43): }\int \frac{1-\cos x}{1+\cos x}\,dx. 
\displaystyle \text{Answer:}
\displaystyle \int\frac{1-\cos x}{1+\cos x}\,dx= \int\frac{(1-\cos x)^2}{1-\cos^2 x}\,dx
\displaystyle = \int\frac{1+\cos^2 x-2\cos x}{\sin^2 x}\,dx
\displaystyle = \int\left(\frac{1}{\sin^2 x}+\frac{\cos^2 x}{\sin^2 x}-\frac{2\cos x}{\sin^2 x}\right)\,dx
\displaystyle = \int\left(\mathrm{cosec}^2 x+\cot^2 x-2\cot x\,\mathrm{cosec}\,x\right)\,dx
\displaystyle = \int\left(2\mathrm{cosec}^2 x-1-2\cot x\,\mathrm{cosec}\,x\right)\,dx
\displaystyle= 2\int \mathrm{cosec}^2 x\,dx-\int dx-2\int \cot x\,\mathrm{cosec}\,x\,dx
\displaystyle = -2\cot x-x+2\mathrm{cosec}\,x+C
\displaystyle = 2(\mathrm{cosec}\,x-\cot x)-x+C
\\

\displaystyle \text{(44): }\int \left\{3\sin x-4\cos x+\frac{5}{\cos^2 x}-\frac{6}{\sin^2 x}+\tan^2 x-\cot^2 x\right\}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(3\sin x-4\cos x+5\sec^2 x-6\mathrm{cosec}^2 x+(\sec^2 x-1)-(\mathrm{cosec}^2 x-1)\right)\,dx
\displaystyle = \int \left(3\sin x-4\cos x+6\sec^2 x-7\mathrm{cosec}^2 x\right)\,dx
\displaystyle = 3\int \sin x\,dx-4\int \cos x\,dx+6\int \sec^2 x\,dx-7\int \mathrm{cosec}^2 x\,dx
\displaystyle = -3\cos x-4\sin x+6\tan x+7\cot x+C
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\displaystyle \textbf{Question 45: }~\text{If }f'(x)=x-\frac{1}{x^2}\text{ and }f(1)=\frac{1}{2},\text{ find }f(x).
\displaystyle \text{Answer:}
\displaystyle f(x)=\int \left(x-\frac{1}{x^2}\right)\,dx
\displaystyle = \int x\,dx-\int x^{-2}\,dx
\displaystyle = \frac{x^{1+1}}{1+1}-\frac{x^{-2+1}}{-2+1}+C
\displaystyle = \frac{x^2}{2}+\frac{1}{x}+C
\displaystyle f(1)=\frac{1}{2}\Rightarrow \frac{1}{2}+1+C=\frac{1}{2}\Rightarrow C=-1
\displaystyle \therefore~f(x)=\frac{x^2}{2}+\frac{1}{x}-1
\\

\displaystyle \textbf{Question 46: }~\text{If }f'(x)=x+b,\;f(1)=5,\;f(2)=13,\text{ find }f(x).
\displaystyle \text{Answer:}
\displaystyle f(x)=\int (x+b)\,dx
\displaystyle = \frac{x^{1+1}}{1+1}+bx+C
\displaystyle = \frac{x^2}{2}+bx+C
\displaystyle f(1)=5\Rightarrow \frac{1}{2}+b+C=5\Rightarrow C=\frac{9}{2}-b
\displaystyle f(2)=13\Rightarrow 2+2b+C=13
\displaystyle 2+2b+\left(\frac{9}{2}-b\right)=13\Rightarrow \frac{13}{2}+b=13\Rightarrow b=\frac{13}{2}
\displaystyle C=\frac{9}{2}-\frac{13}{2}=-2
\displaystyle \therefore~f(x)=\frac{x^2}{2}+\frac{13}{2}x-2
\\

\displaystyle \textbf{Question 47: }~\text{If }f'(x)=8x^3-2x,\;f(2)=8,\text{ find }f(x).
\displaystyle \text{Answer:}
\displaystyle f(x)=\int (8x^3-2x)\,dx
\displaystyle = 8\cdot\frac{x^{3+1}}{3+1}-2\cdot\frac{x^{1+1}}{1+1}+C
\displaystyle = 2x^4-x^2+C
\displaystyle f(2)=8\Rightarrow 2(2^4)-(2^2)+C=8\Rightarrow 32-4+C=8\Rightarrow C=-20
\displaystyle \therefore~f(x)=2x^4-x^2-20
\\

\displaystyle \textbf{Question 48: }~\text{If }f'(x)=a\sin x+b\cos x\text{ and }f'(0)=4,\;f(0)=3,\;f\!\left(\frac{\pi}{2}\right)=5,\text{ find }f(x).
\displaystyle \text{Answer:}
\displaystyle f(x)=\int (a\sin x+b\cos x)\,dx
\displaystyle = -a\cos x+b\sin x+C
\displaystyle f'(0)=4\Rightarrow a\sin 0+b\cos 0=4\Rightarrow b=4
\displaystyle f\!\left(\frac{\pi}{2}\right)=5\Rightarrow -a\cos\frac{\pi}{2}+4\sin\frac{\pi}{2}+C=5\Rightarrow 4+C=5\Rightarrow C=1
\displaystyle f(0)=3\Rightarrow -a\cos 0+4\sin 0+1=3\Rightarrow -a+1=3\Rightarrow a=-2
\displaystyle \therefore~f(x)=2\cos x+4\sin x+1
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\displaystyle \textbf{Question 49: }~\text{Write the primitive or anti-derivative of }f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}.
\displaystyle \text{Answer:}
\displaystyle \int \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\,dx=\int \left(x^{1/2}+x^{-1/2}\right)\,dx
\displaystyle = \frac{x^{1/2+1}}{1/2+1}+\frac{x^{-1/2+1}}{-1/2+1}+C
\displaystyle = \frac{x^{3/2}}{3/2}+\frac{x^{1/2}}{1/2}+C
\displaystyle = \frac{2}{3}x^{3/2}+2\sqrt{x}+C
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