Evaluate $\int{\frac{xe^x}{(1+x)^2} dx}$

Question

How would I evaluate this integral? $$\int{\frac{xe^x}{(1+x)^2} dx}$$ I know I need to use parts but I ended up getting a very complicated expression to integrate the second time.

Answer 1

Let us try integration by parts, with $u=xe^x$ and $dv=\frac{1}{(1+x)^2}\,dx$.

It all collapses nicely. We get $du=(1+x)e^x\,dx$ and we can take $v=-\frac{1}{1+x}$. We end up needing to integrate $e^x$.

Answer 2

$$\int \frac{e^x}{(1+x)^2}dx=-\int e^x d(\frac{1}{x+1})=-\frac{e^x}{x+1}+\int \frac{e^x}{1+x} dx$$

So $$\frac{e^x}{x+1}=\int \frac{e^x}{1+x} dx-\int \frac{e^x}{(1+x)^2}dx=\int \frac{xe^x}{(1+x)^2}dx$$

Answer 3

Same problem. Using integration by parts the integral can be wriren asas

$$ I=\int \frac{xe^x}{(1+x)^2} = \int \frac{e^x}{(1+x)}-\int \frac{e^x}{(1+x)^2}=I_1-I_2. $$

Evaluate $I_2$ using integration by parts gives

$$ I_2 = \int \frac{e^x}{(1+x)^2} = -\frac{e^x}{1+x} + \int\frac{e^x}{1+x}=-\frac{e^x}{1+x}+I_1. $$

Thus the answer is

$$ I= I_1 + \frac{e^x}{1+x}-I_1 = \frac{e^x}{1+x}. $$

Answer 4

Observe that $\displaystyle\frac x{(1+x)^2}=\frac{1+x-1}{(1+x)^2}=\frac1{1+x}-\frac1{(1+x)^2}$

If $\displaystyle f(x)=\frac1{1+x},f'(x)=-\frac1{(1+x)^2}$

Now, $\displaystyle e^x[f(x)+f'(x)]=\frac{d(e^x)}{dx}f(x)+e^x\frac{d[f(x)]}{dx}=\frac{d(e^x\cdot f(x))}{dx}$

$\displaystyle\implies\int e^x[f(x)+f'(x)]dx=\int\frac{d[e^x\cdot f(x)]}{dx}\ dx=\int d[e^x\cdot f(x)]=e^x\cdot f(x)+K$ where $K$ is an arbitrary constant

$\displaystyle\implies\int e^x\cdot\frac x{(1+x)^2}=?$

Answer 5

$$ \begin{aligned} \int \frac{x e^x}{(1+x)^2} d x = & -\int x e^x d\left(\frac{1}{1+x}\right) \\ = & -\frac{x e^x}{1+x}+\int e^x d x \\ = & \frac{e^x}{1+x}+C \end{aligned} $$