Question

Integrate : ex(x2+1/(x+1)2)dx

Answer 1

this is the solution

Answer 2

Given:

             $I &= \int \frac{e^x \left( x^2 + 1 \right)}{\left( x + 1 \right)^2} \,\mathrm dx$

To Find:

• We have to integrate the given function.

Solution:

        Let     $I &= \int \frac{e^x \left( x^2 + 1 \right)}{\left( x + 1 \right)^2} \,\mathrm dx$

Adding and subtracting $2x$

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

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

                 $I &= e^x - 2 \left( \int \frac{e^x}{x + 1} \,\mathrm dx - \int \frac{e^x}{\left( x + 1 \right)^2} \,\mathrm dx \right) \end{align}$

We use integration by parts on the first of these integrals, which uses the following general rule:

                        $[\int u \,\mathrm dv = uv - \int v du\\u = \frac{1}{x+1} \\v =e^x$

   $\therefore \displaystyle \int \frac{e^x}{x+1} \,\mathrm dx = \frac{e^x}{x+1} + \int \frac{e^x}{\left(x+1\right)^2} \,\mathrm dx$

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

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

                        $I&= \boxed{ \frac{e^x\left(x-1\right)}{x+1} + C } \end{align}$

Thus,

                $\int \frac{e^x \left( x^2 + 1 \right)}{\left( x + 1 \right)^2} \,\mathrm dx &= \boxed{ \frac{e^x\left(x-1\right)}{x+1} + C } \end{align}$