# Integrate: e(x^2 +x)(2x+1) dx

1. Jul 1, 2009

### a.a

1. The problem statement, all variables and given/known data

integrate: e(x^2 +x)(2x+1) dx

3. The attempt at a solution

let u= e(x^2 +x)
du=e(x^2 +x)(2x+1)dx

integral e(x^2 +x)(2x+1) dx = integral 1/u du

am I on the right track? i didnt get the same answer as the prof...

2. Jul 1, 2009

Re: integration

If you mean

$$\int e^{x^2+x} (2x+1) \, dx$$

then your substitution is one of at least two that will work, but how do you obtain

$$\int \frac 1 u \, du$$

as the next step?

3. Jul 1, 2009

### a.a

Re: integration

because du=e(x2 +x)(2x+1)dx ?

integral e(x2 +x)(2x+1) dx = integral du/e(x2 +x)
=integral 1/u du
?

4. Jul 1, 2009

Re: integration

If

$$u = e^{x^2+x}$$

then

$$du = e^{x^2+x}(2x+1) dx$$

which is exactly the form of the original integral.
why is there need for a fraction?