integration using generalized power rule

**thedoge** · January 30th 2007, 05:41 PM

integral(x^5*(sqrt(13+x^6))

Here's what I did... but its apparently wrong

1. Realized I needed to use the generalized power rule
2. set x^6+13=u
3. set 6*x^5=du
4. solved for the coeffecient. x^5*Z=du Z=6

so now I have

6*int(u^(1/2))*du

from there i got

(6*u^(3/2))/(3/2)

i plugged in the u...

(6*(x^6+13)^(3/2))/(3/2)

but this is incorrect... what'd i do wrong?

**Soroban** · January 30th 2007, 06:13 PM

Hello, thedoge!

$\int x^5\sqrt{13+x^6}\,dx$

Let $u = 13 + x^6\quad\Rightarrow\quad du = 6x^5dx\quad\Rightarrow\quad dx = \frac{du}{6x^5}$

Substitute: . $\int x^5\cdot u^{\frac{1}{2}} \cdot\frac{du}{6x^5} \;=\;\frac{1}{6}\int u^{\frac{1}{2}}du \;=\;\frac{1}{6}\cdot\frac{2}{3}u^{\frac{3}{2}} + C \;=\;\frac{1}{9}u^{\frac{3}{2}} + C$

Back-substitute: . $\frac{1}{9}(13 + x^6)^{\frac{3}{2}} + C$

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