What is the antiderivative of x/(7x^2+3)^5?

1 Answer
mason m
May 29, 2016

-1/(56(7x^2+3)^4)+C

Explanation:

We wish to find:

intx/(7x^2+3)^5dx

This is a prime case for substitution. Let u=7x^2+3 so that du=14xdx.

Note that we already have xdx in the numerator, so we need to multiply the integrand by 14 to obtain du. Balance this by multiplying the exterior of the integral by 1/14.

=1/14int(14x)/(7x^2+3)^5dx

Now substitute in u and du.

=1/14int(du)/u^5=1/14intu^-5du

Integrating u^-5 results in u^(-5+1)/(-5+1)+C:

=1/14(u^-4/(-4))+C=-1/(56u^4)+C=-1/(56(7x^2+3)^4)+C

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