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"If" int(dx)/((x^(2)-2x+10)^(2))=A("tan"...

`"If" int(dx)/((x^(2)-2x+10)^(2))=A("tan"^(-1)((x-1)/(3))+(f(x))/(x^(2)-2x+10))+C`,where, C is a constant of integration, then

A

`A=1/54` and `f(x) = 9(x-1)^(2)`

B

`A=1/27` and `f(x) =9(x-1)`

C

`A=1/81` and `f(x) = 3(x-1)`

D

`A=1/54` and `f(x) =3(x-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
`int(dx)/(x^(2)-2x+10)^(2)`
Let `x-1=t, dx=dt`
`int(dt)/(t^(2)+9)^(2)`
Now, `int1.(dt)/(t^(2)+9) =1/3 tan^(-1) t/3+c`……………….(1)
Applying integration by parts on L.H.S of (1)
`t/(t^(2)+9) +2 int t^(2)/(t^(2)+9)^(2) dt = 1/3 tan^(-1)t/3 +c`
`t/(t^(2)+9) + 2 int (dt)/(t^(2)+9) - 18int(dt)/(t^(2)+9)^(2) =1/3 tan^(-1) t/3 + c rArr int (dt)/(t^(2)+9)^(2) = 1/54 tan^(-1) tan^(-1) t/3 (3t)/(54(t^(2)+9) +c`
Now put t=x-1
`int(dx)/(x^(2)-2x+10)^(2) = 1/54 (tan^(-1)(x-1)/3 + (3(x-1))/(x^(2) -2x+10))+c`
Hence `A=1/5` and `f(x) = 3(x-1)`
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