Question

$\int \frac{e^x(x^3-x+2)}{{(1+x^2)}^2}dx=$

Answer 1

For this problem we'll use a very interesting result that derivative of $(x+1)e^x$ equals to $(x+2)e^x$.

The given integral can be broken to use above result.

$I=\int e^x\frac{(x^3-x^2+2)}{{(x}^2+1{)}^2}dx$

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

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

Using integration by parts and taking $(x{e}^x+e^x)$ as first and $\frac{2x}{{(x}^2+1{)}^2}$ as second function, we get:

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

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

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

$=e^x\frac{(x+1)}{{(x}^2+1)}+C.$

Hence the given integral equals to $e^x\frac{(x+1)}{{(x}^2+1)}+C.$