# Integration using various techniques

1. Apr 20, 2014

### physics604

1. $$\int e^{x+e^x}\,dx$$

2. Relevant equations

Substitution, integration by parts

3. The attempt at a solution

$$u=e^x$$ $$\int e^{x+e^x}\,dx = \int e^x e^{e^x}\,dx = \int ue^u\,du$$
$$a=u$$ $$da=1du$$ $$dv=e^udu$$ $$v=e^u$$
$$=ue^u-\int e^u\,du = ue^u-e^u$$ $$=e^x e^{e^x}+e^{e^x} = e^{x+e^x}-e^{e^x} + C$$

The textbook's answer was $$e^{e^x} +C$$ How come my answer is different? Any help is much appreciated.

2. Apr 20, 2014

### Staff: Mentor

What is du? It is very important to also write du when you make a substitution u = <something>. In this case (and often), du is NOT equal to dx.

3. Apr 20, 2014

### physics604

$$du=e^xdx$$
How does this work into the calculation?

4. Apr 20, 2014

### haruspex

But that's not what you did. You turned $e^xdx$ into $udu$.

5. Apr 20, 2014

### physics604

∫ex+exdx=∫exeexdx=∫ueudu

Do you mean I should do something like this? $$\int e^x\,dx = \int e^xe^{e^x}\,dx = \int ue^u\,e^x dx$$ But the point of my substitution was to turn everything into u's...

6. Apr 20, 2014

### Staff: Mentor

You have $\int e^x e^{e^x}dx$
u = ex, so du = exdx

What does your integral look like using this substution?

7. Apr 20, 2014

### physics604

Oh, I got it now. Thanks! My integral would be

$$\int e^u\,du$$

8. Apr 20, 2014

Yes.