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#