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

 

                               Integration - By Partial Fraction 

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*Integration of `\int \frac{dx}{x^2 - 4}`

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

                                                

See Also:  

                                                                                                           

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