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

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Answer 1 · 2017-03-06T05:07:33

Please see the explanation.

Write the equation for the expansion:

$4/(x(x^2+4))= A/x+(Bx)/(x^2+4)+C/(x^2+4)$

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

$4= A(x^2+4)+Bx^2+Cx$

Make B and C disappear by letting x = 0:

$4= 4A$

$A = 1$

$4= 1(x^2+4)+Bx^2+Cx$

Let x = 1:

$4= 5 +B+C$

$B + C = -1$

Let x = -1:

$4= 5 +B-C$

$B-C = -1$

Clearly C = 0 and B = -1:

$int4/(x(x^2+4))dx=int1/xdx-intx/(x^2+4)dx$

Modify the second integral for a variable substitution:

$int4/(x(x^2+4))dx=int1/xdx-1/2int(2x)/(x^2+4)dx$

$int4/(x(x^2+4))dx=ln(x)-1/2ln(x^2+4)+C$

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