# Intergration help, reverse chain rule, product rule

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• May 23rd 2009, 09:03 AM
craig
Intergration help, reverse chain rule, product rule
Hi, could someone please verify that this is indeed the correct method for this integral:

$\int x(x^2+8)^2
=
\frac{x}{2x.3} (x^2+8)^3
=
\frac{1}{6} (x^2+8)^3 + C
$
.

If there was a constant outside the brackets then I could have done this no problem, just wasn't too sure what to do with the x outside the brackets.

Thanks for the help.
• May 23rd 2009, 09:25 AM
Spec
It looks correct. You can always differentiate the antiderivative to see if you get the correct answer.
• May 23rd 2009, 12:39 PM
HallsofIvy
Or you could simply multiply it out: $(x^2+ 8)^2= x^4+ 16x^2+ 64$ so $x(x^2+ 8)^2= x^5+ 16x^3+ 64x$ and the integral is $\frac{1}{6}x^6+ 4x^4+ 32x^2+ C$.

$\frac{1}{6}(x^2+ 8)^3+ C= \frac{1}{6}x^6+ \frac{3(8)}{6}x^4+ \frac{3(64)}{6}x^2+ 512+ C$ $= \frac{1}{6}x^6+ 4x^4+ 32x^2+ 512+ C$ which only differs from the previous result by the constant of integration.
• May 25th 2009, 07:15 AM
craig
Thanks for the replys, I should have thought of just differentiating the integral!

Was trying to avoid multiplying out, hoping that my method was right ;)

Thanks again