Question

How do you integrate `int(e^x + 1)^2dx`?

Answer 1

The integral equals `1/2e^(2x) + 2e^x + x + C`

Explanation:

Expand:

`I = int(e^(2x) + 2e^x + 1)dx`

Separate the integrals:

`I = inte^(2x)dx + 2inte^xdx + int 1 dx`

Use `int(e^x) = e^x`:

`I = 1/2e^(2x) + 2e^x + x + C`

Hopefully this helps!

Answer 2

`1/2e^(2x)+2e^x + x + "constant"`

Explanation:

This is a good opportunity to use `u`-substitution. You just might have to use it twice though.

Let `u=e^x` and `du=e^xdx => 1/e^xdu=dx => 1/udu=dx`

So `int(e^x+1)^2dx` becomes `int(u+1)^2/(u)du`

Let `s=u+1` so that `ds=du`

So `int(u+1)^2/(u)du` becomes `ints^2/(s-1)ds`

By long division,

`ints^2/(s-1)ds=int(s+1/(s-1)+1)ds`

Integrate each separately,

`=intsds+int1/(s-1)ds+intds`

`=s^2/2+ln(s-1)+s+"constant"`

Plugging in `s=u+1` gives

`=(u+1)^2/2+ln(u)+u+1+"constant"`

Plugging in `u=e^x` gives

`=(e^x+1)^2/2 + ln(e^x) + e^x + 1 + "constant"`

`=(e^(2x)+2e^x+1)/2 + x + e^x + 1 + "constant"`

`=e^(2x)/2+e^x+1/2 + x + e^x + 1 + "constant"`

`=1/2e^(2x)+2e^x + x + 3/2 + "constant"`

`=1/2e^(2x)+2e^x + x + "constant"` (where `3/2` is absorbed into the constant)