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Trying to solve this tricky one:

$$\int {e^{2x}} \sqrt{1 + e^{2x}}dx$$

I am pretty sure I need to use integration by parts, so I have come up with this so far:

$$u=e^{2x} \Rightarrow du=2e^{2x}$$ $$dv = \sqrt{1+e^{2x}}$$

I am not sure how to integrate to find $v$. Am I on the right track here??

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What about a substitution with $u = 1 + e^{2x}$? – Dylan Moreland Dec 3 '11 at 6:11
Okay, that seems to make more sense. However If I integrate I end up with something messy: $\int e^{2x} \sqrt{1+e^{2x}}dx = (1+e^{2x}) \frac_{e^{2x},2} - \int \frac_{e^{2x},2}2e^{2x}dx$ – Dylan Dec 3 '11 at 6:22
up vote 3 down vote accepted

Let $u=1+e^{2x}$, then $du=2e^{2x}dx$, that is $e^{2x}dx=\frac{1}{2}du$. Using this, the integral can be written as $$\int {e^{2x}} \sqrt{1 + e^{2x}}dx=\int\sqrt{u}\cdot\frac{1}{2}du=\frac{1}{3}u^\frac{3}{2}+C=\frac{1}{3}(1+e^{2x})^\frac{3}{2}+C.$$

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Ah, i see what you did there. Thanks. – Dylan Dec 3 '11 at 7:14
As an aside, it is always difficult to tell whether to use substitution versus integration by parts, etc. – Dylan Dec 3 '11 at 7:22
Alternatively, let $u=\sqrt{1+e^{2x}}$, so $u^2=1+e^{2x}$, $2u\,du=2e^{2x}\,dx$, $e^{2x}\,dx=u\,du$, etc. – Gerry Myerson Dec 3 '11 at 8:38

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