Question

What is the integral of 2xe^x?

Please help me how to solve this problem.

Answer 1

`int2xe^xdx=2(xe^x-e^x)+C`

Explanation:

To find:

`int2xe^xdx`

By constant rule

`int2xe^xdx=2intxe^xdx`

Let `I= intxe^xdx`

`int2xe^xdx=2I`

We can integrate using the parts rule

`intudv=uv-intvdu`

Here,

`u=xtodu=dx`

`dv=e^xdxtov=e^x`

Substituting

`intxe^xdx=xe^x-inte^xdx`

`inte^xdx=e^x`

`intxe^xdx=xe^x-e^x`

`I=xe^x-e^x`

`int2xe^xdx=2(xe^x-e^x)+C`

Answer 2

2xe^(x)-2e^(x)+C

Explanation:

We have: `int(2xe^(x))dx`

This integral can be evaluated using integration by parts.

Let `u=2x Rightarrow frac(du)(dx)=2` and `frac(dv)(dx)=e^(x) Rightarrow v=e^(x)`:

`Rightarrow int(2xe^(x))dx=2xe^(x)-int(2e^(x))dx`

`therefore int(2xe^(x))dx=2xe^(x)-2e^(x)+C`