\displaystyle \textbf{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~\int \frac{1}{4x^2+12x+5}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{dx}{4x^{2}+12x+5}

\displaystyle =\frac{1}{4}\int \frac{dx}{x^{2}+3x+\frac{5}{4}}

\displaystyle =\frac{1}{4}\int \frac{dx}{x^{2}+3x+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+\frac{5}{4}}

\displaystyle =\frac{1}{4}\int \frac{dx}{\left(x+\frac{3}{2}\right)^{2}-\frac{9}{4}+\frac{5}{4}}

\displaystyle =\frac{1}{4}\int \frac{dx}{\left(x+\frac{3}{2}\right)^{2}-1^{2}}

\displaystyle \text{Let } x+\frac{3}{2}=t

\displaystyle \Rightarrow dx=dt

\displaystyle \text{Now, } \frac{1}{4}\int \frac{dx}{\left(x+\frac{3}{2}\right)^{2}-1^{2}}

\displaystyle =\frac{1}{4}\int \frac{dt}{t^{2}-1^{2}}

\displaystyle =\frac{1}{4}\times\frac{1}{2\times1}\log\left|\frac{t-1}{t+1}\right|+C

\displaystyle =\frac{1}{8}\log\left|\frac{x+\frac{3}{2}-1}{x+\frac{3}{2}+1}\right|+C

\displaystyle =\frac{1}{8}\log\left|\frac{x+\frac{1}{2}}{x+\frac{5}{2}}\right|+C

\displaystyle =\frac{1}{8}\log\left|\frac{2x+1}{2x+5}\right|+C

\displaystyle \textbf{Question 2: }~\int \frac{1}{x^2-10x+34}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{dx}{x^{2}-10x+34}

\displaystyle =\int \frac{dx}{x^{2}-10x+25-25+34}

\displaystyle =\int \frac{dx}{(x-5)^{2}+9}

\displaystyle =\int \frac{dx}{(x-5)^{2}+3^{2}}

\displaystyle \text{Let } x-5=t

\displaystyle \Rightarrow dx=dt

\displaystyle \text{Now, } \int \frac{dx}{(x-5)^{2}+3^{2}}

\displaystyle =\int \frac{dt}{t^{2}+3^{2}}

\displaystyle =\frac{1}{3}\tan^{-1}\left(\frac{t}{3}\right)+C

\displaystyle =\frac{1}{3}\tan^{-1}\left(\frac{x-5}{3}\right)+C

\displaystyle \textbf{Question 3: }~\int \frac{1}{1+x-x^2}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{dx}{1+x-x^{2}}

\displaystyle =\int \frac{-dx}{x^{2}-x-1}

\displaystyle =\int \frac{-dx}{x^{2}-x+\frac{1}{4}-\frac{1}{4}-1}

\displaystyle =\int \frac{-dx}{\left(x-\frac{1}{2}\right)^{2}-\frac{5}{4}}

\displaystyle =\int \frac{dx}{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}

\displaystyle =\int \frac{dx}{\left(\frac{\sqrt{5}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}

\displaystyle \text{Let } x-\frac{1}{2}=t

\displaystyle \Rightarrow dx=dt

\displaystyle \text{Now, } \int \frac{dx}{\left(\frac{\sqrt{5}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}

\displaystyle =\int \frac{dt}{\left(\frac{\sqrt{5}}{2}\right)^{2}-t^{2}}

\displaystyle =\frac{1}{2\times\frac{\sqrt{5}}{2}}\log\left|\frac{\frac{\sqrt{5}}{2}+t}{\frac{\sqrt{5}}{2}-t}\right|+C

\displaystyle =\frac{1}{\sqrt{5}}\log\left|\frac{\sqrt{5}+2t}{\sqrt{5}-2t}\right|+C

\displaystyle =\frac{1}{\sqrt{5}}\log\left|\frac{\sqrt{5}+2\left(x-\frac{1}{2}\right)}{\sqrt{5}-2\left(x-\frac{1}{2}\right)}\right|+C

\displaystyle =\frac{1}{\sqrt{5}}\log\left|\frac{\sqrt{5}-1+2x}{\sqrt{5}+1-2x}\right|+C

\displaystyle \textbf{Question 4: }~\int \frac{1}{2x^2-x-1}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{dx}{2x^{2}-x-1}

\displaystyle =\frac{1}{2}\int \frac{dx}{x^{2}-\frac{x}{2}-\frac{1}{2}}

\displaystyle =\frac{1}{2}\int \frac{dx}{x^{2}-\frac{x}{2}+\left(\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}-\frac{1}{2}}

\displaystyle =\frac{1}{2}\int \frac{dx}{\left(x-\frac{1}{4}\right)^{2}-\frac{1}{16}-\frac{1}{2}}

\displaystyle =\frac{1}{2}\int \frac{dx}{\left(x-\frac{1}{4}\right)^{2}-\frac{1+8}{16}}

\displaystyle =\frac{1}{2}\int \frac{dx}{\left(x-\frac{1}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}}

\displaystyle \text{Let } x-\frac{1}{4}=t

\displaystyle \Rightarrow dx=dt

\displaystyle \text{Now, } \frac{1}{2}\int \frac{dx}{\left(x-\frac{1}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}}

\displaystyle =\frac{1}{2}\int \frac{dt}{t^{2}-\left(\frac{3}{4}\right)^{2}}

\displaystyle =\frac{1}{2}\times\frac{1}{2\times\frac{3}{4}}\log\left|\frac{t-\frac{3}{4}}{t+\frac{3}{4}}\right|+C

\displaystyle =\frac{2}{3}\times\frac{1}{2}\log\left|\frac{x-\frac{1}{4}-\frac{3}{4}}{x-\frac{1}{4}+\frac{3}{4}}\right|+C

\displaystyle =\frac{2}{3}\times\frac{1}{2}\log\left|\frac{x-1}{x+\frac{1}{2}}\right|+C

\displaystyle =\frac{1}{3}\log\left|\frac{2(x-1)}{2x+1}\right|+C

\displaystyle =\frac{1}{3}\left[\log\left|\frac{x-1}{2x+1}\right|+\log|2|\right]+C

\displaystyle =\frac{1}{3}\log\left|\frac{x-1}{2x+1}\right|+\frac{1}{3}\log|2|+C

\displaystyle =\frac{1}{3}\log\left|\frac{x-1}{2x+1}\right|+C' \;\;[\because\; C'=\frac{1}{3}\log|2|+C]

\displaystyle \textbf{Question 5: }~\int \frac{1}{x^2+6x+13}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{dx}{x^{2}+6x+13}

\displaystyle =\int \frac{dx}{x^{2}+2\times x\times 3+9-9+13}

\displaystyle =\int \frac{dx}{(x+3)^{2}+2^{2}}

\displaystyle =\frac{1}{2}\tan^{-1}\left(\frac{x+3}{2}\right)+C

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