JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 19)
For real numbers $$\alpha$$, $$\beta$$, $$\gamma$$ and $$\delta $$, if
$$\int {{{({x^2} - 1) + {{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx} $$
$$ = \alpha {\log _e}\left( {{{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \right) + \beta {\tan ^{ - 1}}\left( {{{\gamma ({x^2} + 1)} \over x}} \right) + \delta {\tan ^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right) + C$$
where C is an arbitrary constant, then the value of 10($$\alpha$$ + $$\beta$$$$\gamma$$ + $$\delta$$) is equal to ______________.
$$\int {{{({x^2} - 1) + {{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx} $$
$$ = \alpha {\log _e}\left( {{{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \right) + \beta {\tan ^{ - 1}}\left( {{{\gamma ({x^2} + 1)} \over x}} \right) + \delta {\tan ^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right) + C$$
where C is an arbitrary constant, then the value of 10($$\alpha$$ + $$\beta$$$$\gamma$$ + $$\delta$$) is equal to ______________.
Answer
6
Explanation
$$\int {{{({x^2} - 1) + {{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx} $$
= $$\int {{{{x^2} - 1} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx + \int {{1 \over {{x^4} + 3{x^2} + 1}}dx} } $$
Let, $${I_1} = \int {{{1 - {1 \over {{x^2}}}} \over {\left[ {{{\left( {x + {1 \over x}} \right)}^2} + 1} \right]{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)}}} dx$$
and $${I_2} = \int {{{dx} \over {{x^4} + 3x + 1}}} $$
$${\tan ^{ - 1}}\left( {x + {1 \over x}} \right) = t$$
$${I_1} = \int {{{dt} \over t}} $$
$${I_1} = \ln (t) = \ln \left| {{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)} \right|$$
Now
$${I_2} = \int {{{dx} \over {{x^4} + 3x + 1}}} $$
$$ = {1 \over 2}\int {{{({x^2} + 1) - ({x^2} - 1)} \over {{x^4} + 3{x^2} + 1}}} dx$$
$$ = {1 \over 2}\left[ {\int {{{1 + {1 \over {{x^2}}}} \over {{x^2} + 3 + {1 \over {{x^2}}}}}dx - } \int {{{\left( {1 - {1 \over {{x^2}}}} \right)} \over {{x^2} + 3 + {1 \over {{x^2}}}}}dx} } \right]$$
$$ = {1 \over 2}\left[ {\int {{{1 + {1 \over {{x^2}}}} \over {\left[ {{{\left( {x - {1 \over x}} \right)}^2} + 5} \right]}}} dx - \int {{{1 - {1 \over {{x^2}}}} \over {\left[ {{{\left( {x + {1 \over x}} \right)}^2} + 1} \right]}}} dx} \right]$$
Let $${x - {1 \over x} = u}$$ and $${x + {1 \over x} = v}$$
$$ = {1 \over 2}\left[ {\int {{{du} \over {{u^2} + {{\left( {\sqrt 5 } \right)}^2}}} - \int {{{dv} \over {{v^2} + 1}}} } } \right]$$
$${I_2} = {1 \over {2\sqrt 5 }}{\tan ^{ - 1}}\left( {{{x - {1 \over x}} \over {\sqrt 5 }}} \right) - {1 \over 2}{\tan ^{ - 1}}\left( {x + {1 \over x}} \right)$$
$$I = {I_1} + {I_2} =$$
$$ \ln \left| {{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)} \right| + {1 \over {2\sqrt 5 }}\ln \left( {{{{x^2} - 1} \over {\sqrt 5 x}}} \right) - {1 \over 2}{\tan ^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right) + C$$
$$\alpha = 1,\beta = {1 \over {2\sqrt 5 }},\lambda = {1 \over {\sqrt 5 }},\delta = - {1 \over 2}$$
$$ \therefore $$ $$10(\alpha + \beta \lambda + \delta ) = 10\left[ {1 + {1 \over {10}} - {1 \over 2}} \right]$$
$$ = 10\left( {{1 \over {10}} + {1 \over 2}} \right)$$
$$ = 1 + 5 = 6$$
= $$\int {{{{x^2} - 1} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx + \int {{1 \over {{x^4} + 3{x^2} + 1}}dx} } $$
Let, $${I_1} = \int {{{1 - {1 \over {{x^2}}}} \over {\left[ {{{\left( {x + {1 \over x}} \right)}^2} + 1} \right]{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)}}} dx$$
and $${I_2} = \int {{{dx} \over {{x^4} + 3x + 1}}} $$
$${\tan ^{ - 1}}\left( {x + {1 \over x}} \right) = t$$
$${I_1} = \int {{{dt} \over t}} $$
$${I_1} = \ln (t) = \ln \left| {{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)} \right|$$
Now
$${I_2} = \int {{{dx} \over {{x^4} + 3x + 1}}} $$
$$ = {1 \over 2}\int {{{({x^2} + 1) - ({x^2} - 1)} \over {{x^4} + 3{x^2} + 1}}} dx$$
$$ = {1 \over 2}\left[ {\int {{{1 + {1 \over {{x^2}}}} \over {{x^2} + 3 + {1 \over {{x^2}}}}}dx - } \int {{{\left( {1 - {1 \over {{x^2}}}} \right)} \over {{x^2} + 3 + {1 \over {{x^2}}}}}dx} } \right]$$
$$ = {1 \over 2}\left[ {\int {{{1 + {1 \over {{x^2}}}} \over {\left[ {{{\left( {x - {1 \over x}} \right)}^2} + 5} \right]}}} dx - \int {{{1 - {1 \over {{x^2}}}} \over {\left[ {{{\left( {x + {1 \over x}} \right)}^2} + 1} \right]}}} dx} \right]$$
Let $${x - {1 \over x} = u}$$ and $${x + {1 \over x} = v}$$
$$ = {1 \over 2}\left[ {\int {{{du} \over {{u^2} + {{\left( {\sqrt 5 } \right)}^2}}} - \int {{{dv} \over {{v^2} + 1}}} } } \right]$$
$${I_2} = {1 \over {2\sqrt 5 }}{\tan ^{ - 1}}\left( {{{x - {1 \over x}} \over {\sqrt 5 }}} \right) - {1 \over 2}{\tan ^{ - 1}}\left( {x + {1 \over x}} \right)$$
$$I = {I_1} + {I_2} =$$
$$ \ln \left| {{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)} \right| + {1 \over {2\sqrt 5 }}\ln \left( {{{{x^2} - 1} \over {\sqrt 5 x}}} \right) - {1 \over 2}{\tan ^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right) + C$$
$$\alpha = 1,\beta = {1 \over {2\sqrt 5 }},\lambda = {1 \over {\sqrt 5 }},\delta = - {1 \over 2}$$
$$ \therefore $$ $$10(\alpha + \beta \lambda + \delta ) = 10\left[ {1 + {1 \over {10}} - {1 \over 2}} \right]$$
$$ = 10\left( {{1 \over {10}} + {1 \over 2}} \right)$$
$$ = 1 + 5 = 6$$
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