Is using u-substitution the right approach for integrating e(x^2 + x)(2x+1) dx?

[a.a]

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...

 

[statdad]

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?

 

[a.a]

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

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

 

[statdad]

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?