\displaystyle \text{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~\int \frac{1}{(x-1)\sqrt{x+2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{dx}{(x-1)\sqrt{x+2}}
\displaystyle \text{Putting }x+2=t^2
\displaystyle dx=2t\,dt
\displaystyle I=\int\frac{2t\,dt}{(t^2-3)t}
\displaystyle I=\int\frac{2\,dt}{t^2-3}
\displaystyle I=\int\frac{2\,dt}{t^2-(\sqrt3)^2}
\displaystyle I=\frac{1}{\sqrt3}\log\left|\frac{t-\sqrt3}{t+\sqrt3}\right|+C
\displaystyle I=\frac{1}{\sqrt3}\log\left|\frac{\sqrt{x+2}-\sqrt3}{\sqrt{x+2}+\sqrt3}\right|+C

\displaystyle \textbf{Question 2: }~\int \frac{1}{(x-1)\sqrt{2x+3}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{dx}{(x-1)\sqrt{2x+3}}
\displaystyle \text{Putting }2x+3=t^2
\displaystyle x=\frac{t^2-3}{2}
\displaystyle dx=t\,dt
\displaystyle I=\int\frac{t\,dt}{\left(\frac{t^2-3}{2}-1\right)t}
\displaystyle I=\int\frac{2\,dt}{t^2-5}
\displaystyle I=2\int\frac{dt}{t^2-(\sqrt5)^2}
\displaystyle I=\frac{1}{\sqrt5}\log\left|\frac{t-\sqrt5}{t+\sqrt5}\right|+C
\displaystyle I=\frac{1}{\sqrt5}\log\left|\frac{\sqrt{2x+3}-\sqrt5}{\sqrt{2x+3}+\sqrt5}\right|+C

\displaystyle \textbf{Question 3: }~\int \frac{x+1}{(x-1)\sqrt{x+2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{x+1}{(x-1)\sqrt{x+2}}\,dx
\displaystyle \text{Putting }x+2=t^2
\displaystyle x=t^2-2
\displaystyle dx=2t\,dt
\displaystyle I=\int\frac{(t^2-2+1)2t\,dt}{(t^2-2-1)t}
\displaystyle I=2\int\frac{t^2-1}{t^2-3}\,dt
\displaystyle I=2\int\frac{t^2-3+2}{t^2-3}\,dt
\displaystyle I=2\int dt+4\int\frac{dt}{t^2-3}
\displaystyle I=2t+4\int\frac{dt}{t^2-(\sqrt3)^2}
\displaystyle I=2t+\frac{2}{\sqrt3}\log\left|\frac{t-\sqrt3}{t+\sqrt3}\right|+C
\displaystyle I=2\sqrt{x+2}+\frac{2}{\sqrt3}\log\left|\frac{\sqrt{x+2}-\sqrt3}{\sqrt{x+2}+\sqrt3}\right|+C

\displaystyle \textbf{Question 4: }~\int \frac{x^2}{(x-1)\sqrt{x+2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{x^2}{(x-1)\sqrt{x+2}}\,dx
\displaystyle \text{Putting }x+2=t^2
\displaystyle x=t^2-2
\displaystyle dx=2t\,dt
\displaystyle I=\int\frac{(t^2-2)^2}{(t^2-3)t}\,2t\,dt
\displaystyle I=2\int\frac{(t^2-2)^2}{t^2-3}\,dt
\displaystyle I=2\int\frac{t^4-4t^2+4}{t^2-3}\,dt
\displaystyle \text{Dividing numerator by denominator}
\displaystyle I=2\int\left(t^2-1+\frac{1}{t^2-3}\right)dt
\displaystyle I=2\int(t^2-1)\,dt+2\int\frac{dt}{t^2-3}
\displaystyle I=2\left(\frac{t^3}{3}-t\right)+2\int\frac{dt}{t^2-(\sqrt3)^2}
\displaystyle I=\frac{2t^3}{3}-2t+\frac{1}{\sqrt3}\log\left|\frac{t-\sqrt3}{t+\sqrt3}\right|+C
\displaystyle I=\frac{2}{3}(\sqrt{x+2})^3-2\sqrt{x+2}+\frac{1}{\sqrt3}\log\left|\frac{\sqrt{x+2}-\sqrt3}{\sqrt{x+2}+\sqrt3}\right|+C
\displaystyle I=\frac{2}{3}(x+2)^{3/2}-2\sqrt{x+2}+\frac{1}{\sqrt3}\log\left|\frac{\sqrt{x+2}-\sqrt3}{\sqrt{x+2}+\sqrt3}\right|+C

\displaystyle \textbf{Question 5: }~\int \frac{x}{(x-3)\sqrt{x+1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{x\,dx}{(x-3)\sqrt{x+1}}
\displaystyle \text{Putting }x+1=t^2
\displaystyle x=t^2-1
\displaystyle dx=2t\,dt
\displaystyle I=\int\frac{(t^2-1)2t\,dt}{(t^2-1-3)t}
\displaystyle I=2\int\frac{t^2-1}{t^2-4}\,dt
\displaystyle I=2\int\frac{t^2-4+3}{t^2-4}\,dt
\displaystyle I=2\int dt+6\int\frac{dt}{t^2-4}
\displaystyle I=2t+6\int\frac{dt}{t^2-2^2}
\displaystyle I=2t+\frac{3}{2}\log\left|\frac{t-2}{t+2}\right|+C
\displaystyle I=2\sqrt{x+1}+\frac{3}{2}\log\left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+C

\displaystyle \textbf{Question 6: }~\int \frac{1}{(x^2+1)\sqrt{x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{dx}{(x^2+1)\sqrt{x}}
\displaystyle \text{Putting }x=t^2
\displaystyle dx=2t\,dt
\displaystyle I=\int\frac{2t\,dt}{(t^4+1)t}
\displaystyle I=2\int\frac{dt}{t^4+1}
\displaystyle I=\int\left(\frac{t^2+1}{t^4+1}-\frac{t^2-1}{t^4+1}\right)dt
\displaystyle I=\int\frac{t^2+1}{t^4+1}\,dt-\int\frac{t^2-1}{t^4+1}\,dt
\displaystyle \text{Dividing numerator and denominator by }t^2
\displaystyle I=\int\frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}\,dt-\int\frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}\,dt
\displaystyle I=\int\frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+2}\,dt-\int\frac{1-\frac{1}{t^2}}{\left(t+\frac{1}{t}\right)^2-2}\,dt
\displaystyle \text{Putting }t-\frac{1}{t}=p
\displaystyle \left(1+\frac{1}{t^2}\right)dt=dp
\displaystyle \text{Putting }t+\frac{1}{t}=q
\displaystyle \left(1-\frac{1}{t^2}\right)dt=dq
\displaystyle I=\int\frac{dp}{p^2+(\sqrt2)^2}-\int\frac{dq}{q^2-(\sqrt2)^2}
\displaystyle I=\frac{1}{\sqrt2}\tan^{-1}\left(\frac{p}{\sqrt2}\right)-\frac{1}{2\sqrt2}\log\left|\frac{q-\sqrt2}{q+\sqrt2}\right|+C
\displaystyle I=\frac{1}{\sqrt2}\tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt2}\right)-\frac{1}{2\sqrt2}\log\left|\frac{t+\frac{1}{t}-\sqrt2}{t+\frac{1}{t}+\sqrt2}\right|+C
\displaystyle I=\frac{1}{\sqrt2}\tan^{-1}\left(\frac{t^2-1}{\sqrt2\,t}\right)-\frac{1}{2\sqrt2}\log\left|\frac{t^2-\sqrt2\,t+1}{t^2+\sqrt2\,t+1}\right|+C
\displaystyle I=\frac{1}{\sqrt2}\tan^{-1}\left(\frac{x-1}{\sqrt{2x}}\right)-\frac{1}{2\sqrt2}\log\left|\frac{x-\sqrt{2x}+1}{x+\sqrt{2x}+1}\right|+C

\displaystyle \textbf{Question 7: }~\int \frac{x}{(x^2+2x+2)\sqrt{x+1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{x\,dx}{(x^2+2x+2)\sqrt{x+1}}
\displaystyle I=\int\frac{x\,dx}{\left[(x+1)^2+1\right]\sqrt{x+1}}
\displaystyle \text{Putting }x+1=t^2
\displaystyle x=t^2-1
\displaystyle dx=2t\,dt
\displaystyle I=\int\frac{(t^2-1)2t\,dt}{(t^4+1)t}
\displaystyle I=2\int\frac{t^2-1}{t^4+1}\,dt
\displaystyle \text{Dividing numerator and denominator by }t^2
\displaystyle I=2\int\frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}\,dt
\displaystyle I=2\int\frac{1-\frac{1}{t^2}}{\left(t+\frac{1}{t}\right)^2-2}\,dt
\displaystyle \text{Putting }t+\frac{1}{t}=p
\displaystyle \left(1-\frac{1}{t^2}\right)dt=dp
\displaystyle I=2\int\frac{dp}{p^2-(\sqrt2)^2}
\displaystyle I=\frac{1}{\sqrt2}\log\left|\frac{p-\sqrt2}{p+\sqrt2}\right|+C
\displaystyle I=\frac{1}{\sqrt2}\log\left|\frac{t+\frac{1}{t}-\sqrt2}{t+\frac{1}{t}+\sqrt2}\right|+C
\displaystyle I=\frac{1}{\sqrt2}\log\left|\frac{t^2-\sqrt2\,t+1}{t^2+\sqrt2\,t+1}\right|+C
\displaystyle I=\frac{1}{\sqrt2}\log\left|\frac{x+1-\sqrt{2(x+1)}+1}{x+1+\sqrt{2(x+1)}+1}\right|+C
\displaystyle I=\frac{1}{\sqrt2}\log\left|\frac{x+2-\sqrt{2(x+1)}}{x+2+\sqrt{2(x+1)}}\right|+C

\displaystyle \textbf{Question 8: }~\int \frac{1}{(x-1)\sqrt{x^2+1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{dx}{(x-1)\sqrt{x^2+1}}
\displaystyle \text{Putting }x-1=\frac{1}{t}
\displaystyle dx=-\frac{1}{t^2}\,dt
\displaystyle I=\int\frac{-\frac{1}{t^2}\,dt}{\frac{1}{t}\sqrt{\left(1+\frac{1}{t}\right)^2+1}}
\displaystyle I=\int\frac{-\frac{1}{t}\,dt}{\sqrt{1+\frac{1}{t^2}+\frac{2}{t}+1}}
\displaystyle I=\int\frac{-\frac{1}{t}\,dt}{\sqrt{\frac{2t^2+2t+1}{t^2}}}
\displaystyle I=\int\frac{-dt}{\sqrt{2t^2+2t+1}}
\displaystyle I=-\frac{1}{\sqrt2}\int\frac{dt}{\sqrt{t^2+t+\frac12}}
\displaystyle I=-\frac{1}{\sqrt2}\int\frac{dt}{\sqrt{t^2+t+\frac14-\frac14+\frac12}}
\displaystyle I=-\frac{1}{\sqrt2}\int\frac{dt}{\sqrt{\left(t+\frac12\right)^2+\left(\frac12\right)^2}}
\displaystyle I=-\frac{1}{\sqrt2}\log\left|t+\frac12+\sqrt{\left(t+\frac12\right)^2+\left(\frac12\right)^2}\right|+C
\displaystyle I=-\frac{1}{\sqrt2}\log\left|\frac{1}{x-1}+\frac12+\sqrt{\left(\frac{1}{x-1}+\frac12\right)^2+\frac14}\right|+C

\displaystyle \textbf{Question 9: }~\int \frac{1}{(x+1)\sqrt{x^2+x+1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{dx}{(x+1)\sqrt{x^2+x+1}}
\displaystyle \text{Putting }x+1=\frac{1}{t}
\displaystyle dx=-\frac{1}{t^2}\,dt
\displaystyle I=\int\frac{-\frac{1}{t^2}\,dt}{\frac{1}{t}\sqrt{\left(\frac{1}{t}-1\right)^2+\left(\frac{1}{t}-1\right)+1}}
\displaystyle I=\int\frac{-\frac{1}{t}\,dt}{\sqrt{\frac{1}{t^2}-\frac{1}{t}+1}}
\displaystyle I=-\int\frac{dt}{\sqrt{t^2-t+1}}
\displaystyle I=-\int\frac{dt}{\sqrt{t^2-t+\frac14-\frac14+1}}
\displaystyle I=-\int\frac{dt}{\sqrt{\left(t-\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}}
\displaystyle I=-\log\left|t-\frac12+\sqrt{\left(t-\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}\right|+C
\displaystyle I=-\log\left|\frac{1}{x+1}-\frac12+\sqrt{\left(\frac{1}{x+1}-\frac12\right)^2+\frac34}\right|+C
\displaystyle I=-\log\left|\frac{1}{x+1}-\frac12+\frac{\sqrt{x^2+x+1}}{x+1}\right|+C

\displaystyle \textbf{Question 10: }~\int \frac{1}{(x^2-1)\sqrt{x^2+1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{dx}{(x^2-1)\sqrt{x^2+1}}
\displaystyle \text{Putting }x=\frac{1}{t}
\displaystyle dx=-\frac{1}{t^2}\,dt
\displaystyle I=\int\frac{-\frac{1}{t^2}\,dt}{\left(\frac{1}{t^2}-1\right)\sqrt{\frac{1}{t^2}+1}}
\displaystyle I=\int\frac{-\frac{1}{t^2}\,dt}{\frac{1-t^2}{t^2}\cdot\frac{\sqrt{1+t^2}}{t}}
\displaystyle I=\int\frac{-t\,dt}{(1-t^2)\sqrt{1+t^2}}
\displaystyle \text{Putting }1+t^2=u^2
\displaystyle 2t\,dt=2u\,du
\displaystyle t\,dt=u\,du
\displaystyle I=-\int\frac{u\,du}{(1-(u^2-1))u}
\displaystyle I=-\int\frac{du}{2-u^2}
\displaystyle I=-\int\frac{du}{(\sqrt2)^2-u^2}
\displaystyle I=-\frac{1}{2\sqrt2}\log\left|\frac{u+\sqrt2}{u-\sqrt2}\right|+C
\displaystyle I=-\frac{1}{2\sqrt2}\log\left|\frac{\sqrt{1+t^2}+\sqrt2}{\sqrt{1+t^2}-\sqrt2}\right|+C
\displaystyle I=-\frac{1}{2\sqrt2}\log\left|\frac{\sqrt{1+\frac{1}{x^2}}+\sqrt2}{\sqrt{1+\frac{1}{x^2}}-\sqrt2}\right|+C
\displaystyle I=-\frac{1}{2\sqrt2}\log\left|\frac{\sqrt{x^2+1}+\sqrt2\,x}{\sqrt{x^2+1}-\sqrt2\,x}\right|+C

\displaystyle \textbf{Question 11: }~\int \frac{x}{(x^2+4)\sqrt{x^2+1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{x\,dx}{(x^2+4)\sqrt{x^2+1}}
\displaystyle \text{Putting }x^2=t
\displaystyle 2x\,dx=dt
\displaystyle x\,dx=\frac{dt}{2}
\displaystyle I=\frac12\int\frac{dt}{(t+4)\sqrt{t+1}}
\displaystyle \text{Again putting }t+1=p^2
\displaystyle t=p^2-1
\displaystyle dt=2p\,dp
\displaystyle I=\frac12\int\frac{2p\,dp}{(p^2-1+4)p}
\displaystyle I=\int\frac{dp}{p^2+3}
\displaystyle I=\int\frac{dp}{p^2+(\sqrt3)^2}
\displaystyle I=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{p}{\sqrt3}\right)+C
\displaystyle I=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{\sqrt{t+1}}{\sqrt3}\right)+C
\displaystyle I=\frac{1}{\sqrt3}\tan^{-1}\left(\sqrt{\frac{x^2+1}{3}}\right)+C

\displaystyle \textbf{Question 12: }~\int \frac{1}{(1+x^2)\sqrt{1-x^2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{dx}{(1+x^2)\sqrt{1-x^2}}
\displaystyle \text{Putting }x=\frac{1}{t}
\displaystyle dx=-\frac{1}{t^2}\,dt
\displaystyle I=\int\frac{-\frac{1}{t^2}\,dt}{\left(1+\frac{1}{t^2}\right)\sqrt{1-\frac{1}{t^2}}}
\displaystyle I=\int\frac{-\frac{1}{t^2}\,dt}{\frac{t^2+1}{t^2}\cdot\frac{\sqrt{t^2-1}}{t}}
\displaystyle I=-\int\frac{t\,dt}{(t^2+1)\sqrt{t^2-1}}
\displaystyle \text{Again putting }t^2-1=u^2
\displaystyle 2t\,dt=2u\,du
\displaystyle t\,dt=u\,du
\displaystyle I=-\int\frac{u\,du}{(u^2+1+1)u}
\displaystyle I=-\int\frac{du}{u^2+2}
\displaystyle I=-\int\frac{du}{u^2+(\sqrt2)^2}
\displaystyle I=-\frac{1}{\sqrt2}\tan^{-1}\left(\frac{u}{\sqrt2}\right)+C
\displaystyle I=-\frac{1}{\sqrt2}\tan^{-1}\left(\frac{\sqrt{t^2-1}}{\sqrt2}\right)+C
\displaystyle I=-\frac{1}{\sqrt2}\tan^{-1}\left(\sqrt{\frac{t^2-1}{2}}\right)+C
\displaystyle I=-\frac{1}{\sqrt2}\tan^{-1}\left(\sqrt{\frac{1-x^2}{2x^2}}\right)+C

\displaystyle \textbf{Question 13: }~\int \frac{1}{(2x^2+3)\sqrt{x^2-4}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{dx}{(2x^2+3)\sqrt{x^2-4}}
\displaystyle \text{Putting }x=\frac{1}{t}
\displaystyle dx=-\frac{1}{t^2}\,dt
\displaystyle I=\int\frac{-\frac{1}{t^2}\,dt}{\left(\frac{2}{t^2}+3\right)\sqrt{\frac{1}{t^2}-4}}
\displaystyle I=\int\frac{-\frac{1}{t^2}\,dt}{\frac{2+3t^2}{t^2}\cdot\frac{\sqrt{1-4t^2}}{t}}
\displaystyle I=-\int\frac{t\,dt}{(2+3t^2)\sqrt{1-4t^2}}
\displaystyle \text{Again putting }1-4t^2=u^2
\displaystyle -8t\,dt=2u\,du
\displaystyle t\,dt=-\frac{u}{4}\,du
\displaystyle I=\frac14\int\frac{u\,du}{2+3\left(\frac{1-u^2}{4}\right)}
\displaystyle I=\frac14\int\frac{u\,du}{\frac{11-3u^2}{4}}
\displaystyle I=\int\frac{du}{11-3u^2}
\displaystyle I=\frac13\int\frac{du}{\left(\sqrt{\frac{11}{3}}\right)^2-u^2}
\displaystyle I=\frac{1}{2\sqrt{33}}\log\left|\frac{\sqrt{11}+ \sqrt3\,u}{\sqrt{11}-\sqrt3\,u}\right|+C
\displaystyle I=\frac{1}{2\sqrt{33}}\log\left|\frac{\sqrt{11}+\sqrt3\sqrt{1-4t^2}}{\sqrt{11}-\sqrt3\sqrt{1-4t^2}}\right|+C
\displaystyle I=\frac{1}{2\sqrt{33}}\log\left|\frac{\sqrt{11}+\sqrt{3(1-\frac{4}{x^2})}}{\sqrt{11}-\sqrt{3(1-\frac{4}{x^2})}}\right|+C
\displaystyle I=\frac{1}{2\sqrt{33}}\log\left|\frac{\sqrt{11} x+\sqrt{3x^2-12}}{\sqrt{11} x-\sqrt{3x^2-12}}\right|+C

\displaystyle \textbf{Question 14: }~\int \frac{x}{(x^2+4)\sqrt{x^2+9}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{x\,dx}{(x^2+4)\sqrt{x^2+9}}
\displaystyle \text{Putting }x^2=t
\displaystyle 2x\,dx=dt
\displaystyle x\,dx=\frac{dt}{2}
\displaystyle I=\frac12\int\frac{dt}{(t+4)\sqrt{t+9}}
\displaystyle \text{Again putting }t+9=u^2
\displaystyle t=u^2-9
\displaystyle dt=2u\,du
\displaystyle I=\frac12\int\frac{2u\,du}{(u^2-9+4)u}
\displaystyle I=\int\frac{du}{u^2-5}
\displaystyle I=\int\frac{du}{u^2-(\sqrt5)^2}
\displaystyle I=\frac{1}{2\sqrt5}\log\left|\frac{u-\sqrt5}{u+\sqrt5}\right|+C
\displaystyle I=\frac{1}{2\sqrt5}\log\left|\frac{\sqrt{t+9}-\sqrt5}{\sqrt{t+9}+\sqrt5}\right|+C
\displaystyle I=\frac{1}{2\sqrt5}\log\left|\frac{\sqrt{x^2+9}-\sqrt5}{\sqrt{x^2+9}+\sqrt5}\right|+C

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