Question

$\int \frac{1}{4x^2+4x+5}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{dx}{4x^2+4x+1+4} \\ =\int \frac{dx}{{\left(2x\right)}^2+2\times 2x+1+22} \\ =\int \frac{dx}{{\left(2x+1\right)}^2+2^2} \\ \mathrm{Putting}\left(2x+1\right)=t \\ \Rightarrow 2dx=dt \\ \Rightarrow dx=\frac{dt}{2} \\ \therefore I=\frac{1}{2}\int \frac{dt}{t^2+2^2} \\ =\frac{1}{2}\times \frac{1}{2}{\tan }^{-1}\left(\frac{t}{2}\right)+C \\ =\frac{1}{4}{\tan }^{-1}\left(\frac{2x+1}{2}\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\mathit{\because }\mathit{}t=\left(2x+1\right)\right] \end{gathered}$$