QJEEM19Ma9a5f: JEE Main 2019 Mathematics - Class 12 JEEM19Ma9a5f Solution: If where C is a constant of integration then :

Question

Verified Solution • 2025–26 Session moderate

If

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

where C is a constant of integration then :

Accepted Answer

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

Let{\rm{ }}{\left( {x{\rm{ }}-{\rm{ }}1} \right)^2}{\rm{ }} = {\rm{ }}9ta{n^2}\theta \,\,\,\,...\left( i \right)

\Rightarrow \tan \theta = {{x - 1} \over 3}

On Differentiating ...(i)

2(x – 1)dx = 18tan

\theta

sec²

\theta

d

\theta

\therefore I = \int {{{18\tan \theta {{\sec }^2}\theta d\theta } \over {2 \times 3\tan \theta \times 81{{\sec }^4}\theta }}}

I = {1 \over {27}}\int {{{\cos }^2}\theta \,d\theta } = {1 \over {27}} \times {1 \over 2}\int {\left( {1 + \cos 2\theta } \right)} d\theta

I = {1 \over {54}}\left\{ {\theta + {{\sin 2\theta } \over 2}} \right\} + c

I = {1 \over {54}}\left\{ {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {1 \over 2} \times {{2\left( {{{x - 1} \over 3}} \right)} \over {1 + {{\left( {{{x - 1} \over 3}} \right)}^2}}}} \right\} + c

I = {1 \over {54}}\left[ {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {{3(x - 1)} \over {{x^2} - 2x + 10}}} \right] + c

then

A = {1 \over {54}}

f(x) = 3(x – 1)

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