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If \int \frac{dx}{(x^{2}-2x+10)^{2}}=A(\tan^{-1}(\frac{x-1}{3})+\frac{f(x)}{x^{2}-2x+10})+C

Where C is a constant of integration , then :

  • Option 1)

    A=\frac{1}{54}\: \: and\: \: f(x)=3(x-1)

  • Option 2)

    A=\frac{1}{81}\: \: and\: \: f(x)=3(x-1)

  • Option 3)

    A=\frac{1}{27}\: \: and\: \: f(x)=9(x-1)

  • Option 4)

    A=\frac{1}{54}\: \: and\: \: f(x)=9(x-1)^{2}

 

Answers (1)

\int \frac{dx}{(x^{2}-2x+10)^{2}}=\int \frac{dx}{((x+1)^{2}+9)^{2}}

put x-1=3tan\theta=>dx=3sec^{2}\theta d\theta

I=\int \frac{3sec^{2}\theta d \theta}{81sec^{4}\theta}=\frac{1}{27}\int cos^{2}\theta d \theta

   =\frac{1}{54}\int (1+cos 2\theta) d \theta=\frac{1}{54}(\theta+\frac{sin2\theta}{2})+C

  =\frac{1}{54}[\tan^{-1}(\frac{x-1}{3})+\frac{tan\theta }{sec^{2}\theta }]+C

  =\frac{1}{54}[\tan^{-1}(\frac{x-1}{3})+\frac{\frac{x-1}{3} }{1+(\frac{x-1}{3})^{2} }]+C

 =\frac{1}{54}[\tan^{-1}(\frac{x-1}{3})+\frac{3(x-1)}{x^{2}-2x+10}]+C

A=\frac{1}{54}\: \: and\: \: f(x)=3(x-1)

correct option is (1)

  


Option 1)

A=\frac{1}{54}\: \: and\: \: f(x)=3(x-1)

Option 2)

A=\frac{1}{81}\: \: and\: \: f(x)=3(x-1)

Option 3)

A=\frac{1}{27}\: \: and\: \: f(x)=9(x-1)

Option 4)

A=\frac{1}{54}\: \: and\: \: f(x)=9(x-1)^{2}

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Vakul

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