How do you integrate $\frac{x + 4}{{(x + 1)}^{2} + 4}$ $(x+4)/[(x+1)^2 + 4]$ using partial fractions?

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How do you integrate $\frac{x + 4}{{(x + 1)}^{2} + 4}$ $(x+4)/[(x+1)^2 + 4]$ using partial fractions?

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Answer 1 · 2018-04-18T18:04:49

$\int \frac{x + 4}{{(x + 1)}^{2} + 4} d x = ln \sqrt{{(x + 1)}^{2} + 4} + \frac{3}{2} arctan (\frac{x + 1}{2}) + C$ $int(x+4)/((x+1)^2+4)dx=lnsqrt[(x+1)^2+4]+3/2arctan((x+1)/2)+C$

Explanation:

There is no way to really perform partial fraction decomposition here; however, we may integrate using the following substitution:

$u = x + 1$ $u=x+1$
$x = u - 1$ $x=u-1$
$d u = d x$ $du=dx$

$\int \frac{u - 1 + 4}{u^{2} + 4} d x = \int \frac{u + 3}{u^{2} + 4} d u$ $int(u-1+4)/(u^2+4)dx=int(u+3)/(u^2+4)du$

We may split this up as follows:

$\int \frac{u}{u^{2} + 4} d u + 3 \int \frac{d u}{u^{2} + 4}$ $intu/(u^2+4)du+3int(du)/(u^2+4)$

$\int \frac{u}{u^{2} + 4} d u = \frac{1}{2} ln (u^{2} + 4)$ $intu/(u^2+4)du=1/2ln(u^2+4)$ -- this can be solved with a mental substitution, as the differential of $u^{2} + 4$ $u^2+4$ shows up in the numerator with a bit of simplification. We do not attach absolute value bars as $u^{2} + 4$ $u^2+4$ is always positive.

$3 \int \frac{d u}{u^{2} + 4} = \frac{3}{2} arctan (\frac{u}{2})$ $3int(du)/(u^2+4)=3/2arctan(u/2)$ .

This comes from the common integral, $\int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} arctan (\frac{x}{a})$ $intdx/(x^2+a^2)=1/aarctan(x/a)$ . For our case, $a = \sqrt{4} = 2$ $a=sqrt4=2$ .

Thus, rewriting in terms of $x$ $x$ and simplifying the logarithm, we see

$\int \frac{x + 4}{{(x + 1)}^{2} + 4} d x = ln \sqrt{{(x + 1)}^{2} + 4} + \frac{3}{2} arctan (\frac{x + 1}{2}) + C$ $int(x+4)/((x+1)^2+4)dx=lnsqrt[(x+1)^2+4]+3/2arctan((x+1)/2)+C$

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How do you integrate $\frac{x + 4}{{(x + 1)}^{2} + 4}$ $(x+4)/[(x+1)^2 + 4]$ using partial fractions?

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How do you integrate (x+4)/[(x+1)^2 + 4]x+4(x+1)2+4(x+4)/[(x+1)^2 + 4] using partial fractions?

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How do you integrate $\frac{x + 4}{{(x + 1)}^{2} + 4}$ $(x+4)/[(x+1)^2 + 4]$ using partial fractions?