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# Indefinite Integrals Using U-Substitution

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I get integrals and indefinite integrals. I do. I just don't get this one:

$$\int\frac{e^{sqrt(x)}}{sqrt(x)}$$

By the way, I couldn't figure out how to get square roots using LaTeX, so sqrt is square root.

ThisGuy  Feb 24, 2015

#1
+26753
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1. Use \sqrt{x} to get  $$\sqrt{x}$$

2. Use the substitution:

$$u=\sqrt{x}$$

$$\\du=\frac{dx}{2\sqrt{x}}\\2\sqrt{x}du=dx\\2udu=dx$$

So:

$$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=\int \frac{e^{u} \times 2udu}{u}$$

or:

$$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int e^{u}du = 2e^u=2e^{\sqrt{x}}$$

.

Alan  Feb 24, 2015
#1
+26753
+10

1. Use \sqrt{x} to get  $$\sqrt{x}$$

2. Use the substitution:

$$u=\sqrt{x}$$

$$\\du=\frac{dx}{2\sqrt{x}}\\2\sqrt{x}du=dx\\2udu=dx$$

So:

$$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=\int \frac{e^{u} \times 2udu}{u}$$

or:

$$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int e^{u}du = 2e^u=2e^{\sqrt{x}}$$

.

Alan  Feb 24, 2015
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+227
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Ugh, now this seems way too simple. Thank you a ton!

ThisGuy  Feb 24, 2015