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

## Homework Statement

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

## 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...

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Homework Helper

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?

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

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

Homework Helper

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?