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

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

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Answer 1 · 2017-03-06T05:07:33

Please see the explanation.

Explanation:

Write the equation for the expansion:

$\frac{4}{x (x^{2} + 4)} = \frac{A}{x} + \frac{B x}{x^{2} + 4} + \frac{C}{x^{2} + 4}$ $4/(x(x^2+4))= A/x+(Bx)/(x^2+4)+C/(x^2+4)$

Multiply both sides by $x (x^{2} + 4)$ $x(x^2+4)$ :

$4 = A (x^{2} + 4) + B x^{2} + C x$ $4= A(x^2+4)+Bx^2+Cx$

Make B and C disappear by letting x = 0:

$4 = 4 A$ $4= 4A$

$A = 1$ $A = 1$

$4 = 1 (x^{2} + 4) + B x^{2} + C x$ $4= 1(x^2+4)+Bx^2+Cx$

Let x = 1:

$4 = 5 + B + C$ $4= 5 +B+C$

$B + C = - 1$ $B + C = -1$

Let x = -1:

$4 = 5 + B - C$ $4= 5 +B-C$

$B - C = - 1$ $B-C = -1$

Clearly C = 0 and B = -1:

$\int \frac{4}{x (x^{2} + 4)} d x = \int \frac{1}{x} d x - \int \frac{x}{x^{2} + 4} d x$ $int4/(x(x^2+4))dx=int1/xdx-intx/(x^2+4)dx$

Modify the second integral for a variable substitution:

$\int \frac{4}{x (x^{2} + 4)} d x = \int \frac{1}{x} d x - \frac{1}{2} \int \frac{2 x}{x^{2} + 4} d x$ $int4/(x(x^2+4))dx=int1/xdx-1/2int(2x)/(x^2+4)dx$

$\int \frac{4}{x (x^{2} + 4)} d x = ln (x) - \frac{1}{2} ln (x^{2} + 4) + C$ $int4/(x(x^2+4))dx=ln(x)-1/2ln(x^2+4)+C$

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

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How do you integrate 4(x(x2+4))4/((x(x^2+4)) using partial fractions?

1 Answer

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