Verified Solution • 2025–26 Session moderate

QJEEM21Ma6037: JEE Main 2021 Mathematics - Class 12 JEEM21Ma6037 Solution: For real numbers , , and , if where C is an arbitrary constant, …

For real numbers α\alpha , β\beta , γ\gamma and δ\delta , if
∫(x2−1)+tan⁡−1(x2+1x)(x4+3x2+1)tan⁡−1(x2+1x)dx\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}

=αlog⁡e(tan⁡−1(x2+1x))+βtan⁡−1(γ(x2+1)x)+δtan⁡−1(x2+1x)+C= \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( α\alpha + β\beta γ\gamma + δ\delta ) is equal to ______________.

Step-by-Step Verified Solution

∫(x2−1)+tan⁡−1(x2+1x)(x4+3x2+1)tan⁡−1(x2+1x)dx\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}

= ∫x2−1(x4+3x2+1)tan⁡−1(x2+1x)dx+∫1x4+3x2+1dx\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, I1=∫1−1x2[(x+1x)2+1]tan⁡−1(x+1x)dx{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 I2=∫dxx4+3x+1{I_2} = \int {{{dx} \over {{x^4} + 3x + 1}}}

tan⁡−1(x+1x)=t{\tan ^{ - 1}}\left( {x + {1 \over x}} \right) = t

I1=∫dtt{I_1} = \int {{{dt} \over t}}

I1=ln⁡(t)=ln⁡∣tan⁡−1(x+1x)∣{I_1} = \ln (t) = \ln \left| {{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)} \right|

Now

I2=∫dxx4+3x+1{I_2} = \int {{{dx} \over {{x^4} + 3x + 1}}}

=12∫(x2+1)−(x2−1)x4+3x2+1dx= {1 \over 2}\int {{{({x^2} + 1) - ({x^2} - 1)} \over {{x^4} + 3{x^2} + 1}}} dx

=12[∫1+1x2x2+3+1x2dx−∫(1−1x2)x2+3+1x2dx]= {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]

=12[∫1+1x2[(x−1x)2+5]dx−∫1−1x2[(x+1x)2+1]dx]= {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−1x=u{x - {1 \over x} = u} and x+1x=v{x + {1 \over x} = v}

=12[∫duu2+(5)2−∫dvv2+1]= {1 \over 2}\left[ {\int {{{du} \over {{u^2} + {{\left( {\sqrt 5 } \right)}^2}}} - \int {{{dv} \over {{v^2} + 1}}} } } \right]

I2=125tan⁡−1(x−1x5)−12tan⁡−1(x+1x){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=I1+I2=I = {I_1} + {I_2} =

ln⁡∣tan⁡−1(x+1x)∣+125ln⁡(x2−15x)−12tan⁡−1(x2+1x)+C\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

α=1,β=125,λ=15,δ=−12\alpha = 1,\beta = {1 \over {2\sqrt 5 }},\lambda = {1 \over {\sqrt 5 }},\delta = - {1 \over 2}

∴\therefore 10(α+βλ+δ)=10[1+110−12]10(\alpha + \beta \lambda + \delta ) = 10\left[ {1 + {1 \over {10}} - {1 \over 2}} \right]

=10(110+12)= 10\left( {{1 \over {10}} + {1 \over 2}} \right)

=1+5=6= 1 + 5 = 6

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