# How do I integrate `int e^{2x}(5-x)dx` ? I'm still not too sure on how to integrate by parts.

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### 1 Answer

This integral can be divided into two integrals. Consider:

`int e^{2x}5dx` let `u=2x` , then `du=2dx`

`=5/2inte^udu`

`=5/2e^u+C_1` where `C_1` is a constant of integration.

`=5/2e^{2x}+C_1`

Now the second integral becomes:

`-int xe^{2x}dx` use integration by parts formula `int udv=uv-int vdu` where `u=x` and `dv=e^{2x}dx` . This means `du=dx` and `v=1/2e^{2x}` from the previous integral.

So the second integral becomes:

`-1/2xe^{2x}+int 1/2 e^{2x}dx`

`=-1/2xe^{2x}+1/4e^{2x}+C_2` where C_2 is a constant of integration

**So the original integral is `5/2e^{2x}-1/2xe^{2x}+1/4e^{2x}+C=11/4e^{2x}-1/2xe^{2x}+C` , where `C=C_1+C_2` is a constant of integration.**