How do you integrate $\int \frac{x^{2} + 4}{x + 2}$ $int (x^2+4)/(x+2)$ using substitution?

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Answer 1 · 2016-10-02T20:50:55

$\frac{x^{2} - 4 x + 16 ln | x + 2 |}{2} + C$ $(x^2-4x+16lnabs(x+2))/2+C$

$I = \int \frac{x^{2} + 4}{x + 2} d x$ $I=int(x^2+4)/(x+2)dx$

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

$I = \int \frac{x^{2}}{x + 2} d x + \int \frac{4}{x + 2} d x$ $I=intx^2/(x+2)dx+int4/(x+2)dx$

Perform long division on $x^{2} \div (x + 2)$ $x^2-:(x+2)$ to see that $\frac{x^{2}}{x + 2} = x - 2 + \frac{4}{x + 2}$ $x^2/(x+2)=x-2+4/(x+2)$ :

$I = \int x d x - 2 \int d x + 8 \int \frac{d x}{x + 2}$ $I=intxdx-2intdx+8intdx/(x+2)$

The first two don't require substitution:

$I = \frac{x^{2}}{2} - 2 x + 8 \int \frac{d x}{x + 2}$ $I=x^2/2-2x+8intdx/(x+2)$

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

$I = \frac{x^{2} - 4 x}{2} + 8 \int \frac{d u}{u}$ $I=(x^2-4x)/2+8int(du)/u$

This is a common integral:

$I = \frac{x^{2} - 4 x}{2} + 8 ln | u | + C$ $I=(x^2-4x)/2+8lnabsu+C$

Back-substitute $u = x + 2$ $u=x+2$ :

$I = \frac{x^{2} - 4 x + 16 ln | x + 2 |}{2} + C$ $I=(x^2-4x+16lnabs(x+2))/2+C$

How do you integrate ∫x2+4x+2int (x^2+4)/(x+2) using substitution?