Evaluate : `int (x^(2))/((x^(2) -1)(x^(2) +2)) dx`

Options

A. `1/6 log |(5x-1)/(x+1)| +(sqrt(2))/(3) tan^(-1).(x)/(sqrt(2)) +C`
B. `1/6 log |(x-1)/(x+1)| +(sqrt(2))/(3) tan^(-1).(x)/(sqrt(2)) +C`
C. `1/6 log |(x-1)/(7x+1)| +(sqrt(2))/(3) tan^(-1).(x)/(sqrt(2)) +C`
D. `1/6 log |(2x-1)/(x+1)| +(sqrt(2))/(3) tan^(-1).(x)/(sqrt(2)) +C`

Answer

B

Explanation

`int (x^(2))/((x^(2)-1)(x^(2)+2)) dx`
`"Now " (x^(2))/((x^(2)-1)(x^(2)+2)) =(t)/((t-1)(t+2))`
`"(Where "x^(2) "=t)"`
`=(A)/(t-1)+(B)/(t+2)`
` =(A(t+2)+B(t-1))/((t-1)(t+2))`
`rArr A(t+2)+B(t-1) =t`
Comparing the coefficients of t and constant terms
`A +B =1`
`2A-B=0`
On solving
`A =(1)/(3),B =(2)/(3)`
`:. (t)/((t-1)(t+2))=(1)/(3(t-1))+(2)/(3(t-1))`
`rArr (x^(2))/((x^(2)-1)(x^(2)+2))=(1)/(3(x^(2)-1))+(2)/(3(x^(2)+2))`
`rArr int (x^(2))/((x^(2)+1)(x^(2)+2))dx`
` =1/3 int (1)/(x^(2)-1) dx +2/3 int(1)/(x^(2)+2) dx`
`=1/3 .1/2 log |(x-1)/(x+1)| +2/3 .1/(sqrt(2)) tan^(-1) .(x)/(sqrt(2)) +c`
` =1/6 log |(x-1)/(x+1)| +(sqrt(2))/(3) tan^(-1).(x)/(sqrt(2)) +C.`