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Calculus, EXT2 C1 EQ-Bank 18

Using partial fractions, show that

`int_0^(1/2) 8/(1-x^4)\ dx = log_e 9-4 tan^(−1) (1/2)`  (3 marks)

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`text(Using partial fractions:)`

`8/(1-x^4) = A/(1-x^2) + B/(1 + x^2)`

`A(1 + x^2) + B(1-x^2)` `= 8`
`A + B + (A-B)x^2` `= 8`
`A + B` ` = 8\ \ …\ (1)`
`A-B` ` = 0\ \ …\ (2)`

 
`A = 4, \ B = 4`

 

`4/(1-x^2) = A/(1-x) + B/(1 + x)`

`A(1 + x) + B(1-x)` `= 4`
`A + B + (A-B)x` `= 4`
`A + B` `= 4\ \ …\ (1)`
`A-B` `= 0\ \ …\ (2)`

 
`A = 2, B = 2`
 

`int_0^(1/2) 8/(1-x^4)\ dx` `= int_0^(1/2) 2/(1-x) + 2/(1 + x) + 4/(1 + x^2)\ dx`
  `= [−2ln |1-x| + 2ln |1 + x| + 4tan^(−1)x]_0^(1/2)`
  `= [2ln |(1 + x)/(1-x)| + 4tan^(−1)x]_0^(1/2)`
  `= 2ln |(1 1/2)/(1/2)| + 4tan^(−1)(1/2)-(2ln1 + 4tan^(−1) 0)`
  `= 2ln3 + 4tan^(−1)(1/2)`
  `= ln9 + 4tan^(−1)(1/2)`