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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)/(27)andf(x)=9(x-1)`

B

`A=(1)/(81)andf(x)=3(x-1)`

C

`A=(1)/(54)andf(x)=3(x-1)`

D

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

Text Solution

Verified by Experts

The correct Answer is:
C
Let `I+int(dx)/((x^(2)-2x+10)^(2))=int(dx)/(((x-1)^(2)+3^(2))^(2))` Now, put `x-1 = 3 tan theta rArr dx = 3 sec^(2)theta d theta`
So, `I = int(3sec^(2)theta d theta)/((3^(2)tan^(2)theta+3^(2))^(2))=int(3sec^(2)theta d theta )/(3^(4)sec^(4)theta)`
`=(1)/(27)intcos^(2)theta d theta = (1)/(27)int(1+cos 2 theta)/(2)d theta`
`" "[therefore cos^(2)theta=(1+cos2theta)/(2)]`
`=(1)/(54)int(1+cos 2 theta)d theta = (1)/(54)(theta+(sin 2 theta)/(2))+C`
`=(1)/(54)tan^(-1)((x-1)/(3))+(1)/(108)((2 tan theta)/(1 + tan^(2)theta))+C`
`" "[therefore sin 2 theta = (2 tan theta)/(1+tan^(2)theta)]`
`=(1)/(54)tan^(-1)((x-1)/(3))+(1)/(54)(((x-1)/(3)))/(1+((x-1)/(3))^(2))+C`
`=(1)/(54)tan^(-1)((x-1)/(3))+(1)/(18)((x-1)/((x-1)^(2)+3^(2)))+C`
`=(1)/(54)tan^(-1)((x-1)/(3))+(1)/(18)((x-1)/(x^(2)-2x+10))+C`
`=(1)/(54)[tan^(-1)((x-1)/(3))+(3(x-1))/(x^(2)-2x + 10)]+C`
It is given, that
`I=A[tan^(-1)((x-1)/(3))+(f(x))/(x^(2)-2x + 10)]+C`
On comparing, we get `A=(1)/(54)and f(x) = 3(x-1)`.
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