Question

$\int \frac{x^2+3x+1}{{\left(x+1\right)}^2}dx$

Answer 1

$$\begin{gathered} \int \left(\frac{x^2+3x+1}{{\left(x+1\right)}^2}\right)dx \\ \mathrm{Let}x+1=t \\ \Rightarrow x=t-1 \\ \Rightarrow 1=\frac{dt}{dx} \\ \Rightarrow dx=dt \\ \mathrm{Now},\int \left(\frac{x^2+3x+1}{{\left(x+1\right)}^2}\right)dx \\ =\int \left[\frac{{\left(t-1\right)}^2+3\left(t-1\right)+1}{t^2}\right]dt \\ =\int \left(\frac{t^2-2t+1+3t-3+1}{t^2}\right)dt \\ =\int \left(\frac{t^2+t-1}{t^2}\right)dt \\ =\int \left(1+\frac{1}{t}-t^{-2}\right)dt \\ =t+\log \left|t\right|-\frac{t^{-2+1}}{-2+1}+\mathrm{C} \\ =t+\log \left|t\right|+\frac{1}{t}+\mathrm{C} \\ =x+1+\log \left|x+1\right|+\frac{1}{x+1}+\mathrm{C} \\ \mathrm{Let}1+\mathrm{C}=\mathrm{C}\text{'} \\ =x+\log \left|x+1\right|+\frac{1}{x+1}+\mathrm{C}\text{'} \end{gathered}$$