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 $u=x+2$ :

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

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