integration by partial fraction | integral of `\int \frac{dx}{x^2 - 4}` | Integration Solved Examples

Integration - By Partial Fraction

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Sol: Given I = `\int \frac{dx}{x^2 - 4}`

Let ` \frac{1}{x^2 - 4}` = ` \frac{1}{(x - 2)(x +2)}`

Now decomposing in partial fractions

` \frac{1}{(x - 2)(x +2)}` = ` \frac{A}{x - 2}` +` \frac{B}{x +2}` .....................(1)

= ` \frac{A.(x+2) + B.(x-2)}{x^2 - 4}`

OR = ` \frac{(A+B).x +(2A-2B)}{x^2 - 4}`

Now Comparing coefficients of 'x' and constants on both sides of equal sign, we get

A+B = 0 i.e., A = -B ...............(2)

& 2A - 2B = 1 ............(3)

Now putting the value of ' A' from equation (2) to equation (3) we get.

-2B - 2B = 1

OR -4B = 1 => B = - ` \frac{1}{4}`

& from the value of 'B ' we get 'A' as: A = ` \frac{1}{4}`

Now substituting these values in equation (1) we get.

` \frac{1}{(x - 2)(x +2)}` = `\frac{1}{4} \frac{1}{x - 2}` - ` \frac{1}{4} \frac{1}{x +2}`

Now performing integration on both sides w.r.t x.

= `\frac{1}{4}\int \frac{dx}{x - 2}` - ` \frac{1}{4} \int\frac{dx}{x +2}`

= `\frac{1}{4}.log(x - 2)` - ` \frac{1}{4}.log(x + 2)` + c

OR I = `\frac{1}{4}.log\frac{x - 2}{x + 2} ` + c

$integral of `\int \frac{dx}{x^2 - 4}`$