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Integrate $\int {e}^{3\sqrt x} dx$ ...

let $u = \sqrt x, du = \frac {1}{2\sqrt x}, 2u du = dx$

Then I used integration by parts:

$\int f dg = fg - \int gdf \\ f = u; df = du; \\ dg = e^{3u} du ;g = \int e^{3u} du = \frac 13 e^{3u}\\ \int f dg = fg - \int gdf \Rightarrow \frac 13 e^{3u}u - \frac 13 \int e^{3u}du = \frac 13 e^{3u} (u- \frac 13) + c \Rightarrow \frac 13 e^{3\sqrt x}(\sqrt x - \frac 13) + C $

My question is that why is my answer different from the given answer

$\frac 29 e^{3\sqrt x}(3 \sqrt x -1)$

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Factor out the 3. –  Stefan Nov 14 '12 at 17:17
It isn't different –  Hagen von Eitzen Nov 14 '12 at 17:20
While doing integration by parts, u forgot to multiply the 2 you got from $2udu$. –  Gautam Shenoy Nov 14 '12 at 17:20
oh gosh. thanks for pointing out my careless mistake.... –  melyong Nov 14 '12 at 17:28
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1 Answer 1

up vote 1 down vote accepted

First, notice that

$$\frac{1}{3}e^{3\sqrt{x}}\left(\sqrt{x}-\frac{1}{3}\right) = \frac{1}{9}e^{3\sqrt{x}}\left(3\sqrt{x}-1\right).$$

You can multiply the last bracket by $3$ as long as you divide everything by $3$ to compensate. It seems that your answer is exactly half of what the book is suggesting.

During the substitution step, you had $u = \sqrt{x}$ and $dx = 2u \, du$. It follows that

$$\int e^{3\sqrt{x}} \, dx \equiv \int 2ue^{3u} \, du \, . $$

However, when you integrate by parts you put $f = u$ and $dg = e^{3u} \, du$ when, in fact you really need to put $f = 2u$ and $dg = e^{3u} \, du$. This leads to $df = 2 \, du$ and $g = \frac{1}{3}e^{3u}.$ Putting the pieces together:

$$\int 2ue^{3u} \, du = \frac{2}{3}ue^{3u} - \int \frac{2}{3}e^{3u} \, du \, .$$

This is in fact double what you got, and would lead to the correct solution.

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thank you so much for your clear and detailed explanation. I really appreciate your help.. :D –  melyong Nov 14 '12 at 17:40
@melyong My pleasure. I'm glad I was able to help. –  Fly by Night Nov 14 '12 at 17:56
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