Mathematics - Indefinite Integration Question with Solution | TestHub

The given integral is $I=\int e^x\frac{x^2+1}{(x+1{)}^2}dx$. We can rewrite the rational function part:$\frac{x^2+1}{(x+1{)}^2}=\frac{x^2+2x+1-2x}{(x+1{)}^2}=1-\frac{2x}{(x+1{)}^2}$. Further manipulation: $1-\frac{2(x+1-1)}{(x+1{)}^2}=1-\frac{2}{x+1}+\frac{2}{(x+1{)}^2}$. So, $I=\int e^x\left(1-\frac{2}{x+1}+\frac{2}{(x+1{)}^2}\right)dx$. Let $f(x)=\frac{x-1}{x+1}$. Then $f^{\prime }(x)=\frac{(1)(x+1)-(x-1)(1)}{(x+1{)}^2}=\frac{2}{(x+1{)}^2}$. We notice that $1-\frac{2}{x+1}=\frac{x+1-2}{x+1}=\frac{x-1}{x+1}$. So the integral is of the form $\int e^x(f(x)+f^{\prime }(x))dx=e^{x}f(x)+C$. Thus, $I=e^x\frac{x-1}{x+1}+C$.