How do you integrate # x^3((x^4)+3)^5 dx#?

1 Answer
mason m
May 3, 2016

#(x^4+3)^6/24+C#

Explanation:

We have the integral

#intx^3(x^4+3)^5dx#

This can be solved fairly simply through substitution: we don't have to go through the hassle of expanding the binomial and then integrating term by term.

Let #u=x^4+3#. This implies that #du=4x^3dx#. Since we only have #x^3dx# in the integral and not #4x^3dx#, multiply the interior of the integral by #4# and the exterior by #1//4# to balance this.

#intx^3(x^4+3)^5dx=1/4int(x^4+3)^5*4x^3dx=1/4intu^5du#

Integrating this results in #(1/4)u^6/6+C#, and since #u=x^4+3#, this becomes #(x^4+3)^6/24+C#.

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