How do you integrate $\int \frac{x + 4}{x^{2} + 2 x + 5} d x$ $int (x+4)/(x^2 + 2x + 5)dx$ using partial fractions?

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How do you integrate $\int \frac{x + 4}{x^{2} + 2 x + 5} d x$ $int (x+4)/(x^2 + 2x + 5)dx$ using partial fractions?

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Answer 1 · 2016-04-01T02:00:40

$\frac{1}{2} [ln ∣ ∣ x^{2} + 2 x + 5 ∣ ∣ + 3 arctan (\frac{x + 1}{2})] + c$ $1/2[ln|x^2+2x+5|+3arctan((x+1)/2)]+c$

Explanation:

First split the integral into two parts:

$x + 1$ $x+1$ is a scalable factor of the derivative of $x^{2} + 2 x + 5$ $x^2+2x+5$ , so divide $x + 4$ $x+4$ by $x + 1$ $x+1$

$\int \frac{x + 4}{x^{2} + 2 x + 5} d x = \int \frac{x + 1}{x^{2} + 2 x + 5} d x + \int \frac{3}{x^{2} + 2 x + 5} d x$ $int(x+4)/(x^2+2x+5)dx=int(x+1)/(x^2+2x+5)dx+int3/(x^2+2x+5)dx$

Algebraic manipulation yields:
$\frac{1}{2} \int \frac{2 x + 2}{x^{2} + 2 x + 5} d x + 3 \int \frac{1}{{(x + 1)}^{2} + 4} d x$ $1/2int(2x+2)/(x^2+2x+5)dx+3int1/((x+1)^2+4)dx$
$= \frac{1}{2} ln ∣ ∣ x^{2} + 2 x + 5 ∣ ∣ + \frac{3}{2} arctan (\frac{x + 1}{2}) + c$ $=1/2ln|x^2+2x+5|+3/2arctan((x+1)/2)+c$
$= \frac{1}{2} [ln ∣ ∣ x^{2} + 2 x + 5 ∣ ∣ + 3 arctan (\frac{x + 1}{2})] + c$ $=1/2[ln|x^2+2x+5|+3arctan((x+1)/2)]+c$

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How do you integrate $\int \frac{x + 4}{x^{2} + 2 x + 5} d x$ $int (x+4)/(x^2 + 2x + 5)dx$ using partial fractions?

1 Answer

Explanation:

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How do you integrate int (x+4)/(x^2 + 2x + 5)dx∫x+4x2+2x+5dxint (x+4)/(x^2 + 2x + 5)dx using partial fractions?

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How do you integrate $\int \frac{x + 4}{x^{2} + 2 x + 5} d x$ $int (x+4)/(x^2 + 2x + 5)dx$ using partial fractions?