JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 16)
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 :
where C is a constant of integration then :
A =$${1 \over {54}}$$ and f(x) = 9(x–1)2
A =$${1 \over {54}}$$ and f(x) = 3(x–1)
A =$${1 \over {81}}$$ and f(x) = 3(x–1)
A =$${1 \over {27}}$$ and f(x) = 9(x–1)2
Explanation
$$\int {{{dx} \over {{{({x^2} - 2x + 10)}^2}}} = \int {{{dx} \over {{{({{(x - 1)}^2} + 9)}^2}}}} } $$
$$Let{\rm{ }}{\left( {x{\rm{ }}-{\rm{ }}1} \right)^2}{\rm{ }} = {\rm{ }}9ta{n^2}\theta \,\,\,\,...\left( i \right)$$
$$ \Rightarrow \tan \theta = {{x - 1} \over 3}$$
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)
$$Let{\rm{ }}{\left( {x{\rm{ }}-{\rm{ }}1} \right)^2}{\rm{ }} = {\rm{ }}9ta{n^2}\theta \,\,\,\,...\left( i \right)$$
$$ \Rightarrow \tan \theta = {{x - 1} \over 3}$$
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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