How do you integrate #int (x^2+4)/(x+2)# using substitution?

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Answer 1 · 2016-10-02T20:50:55

#(x^2-4x+16lnabs(x+2))/2+C#

#I=int(x^2+4)/(x+2)dx#

Before performing any integration, let's rewrite this function:

#I=intx^2/(x+2)dx+int4/(x+2)dx#

Perform long division on #x^2-:(x+2)# to see that #x^2/(x+2)=x-2+4/(x+2)#:

#I=intxdx-2intdx+8intdx/(x+2)#

The first two don't require substitution:

#I=x^2/2-2x+8intdx/(x+2)#

For the final integral, let #u=x+2#, so #du=dx#:

#I=(x^2-4x)/2+8int(du)/u#

This is a common integral:

#I=(x^2-4x)/2+8lnabsu+C#

Back-substitute #u=x+2#:

#I=(x^2-4x+16lnabs(x+2))/2+C#