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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