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Joint Entrance Examination
Mathematics
Indefinite Integrals
Question 373
Question 373 of 416
Medium
MCQ
The integral \[\int {{{2{x^{12}} + 5{x^9}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx\] is equal to :
A
\[{{{x^5}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C\]
B
\[{{ - {x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C\]
C
\[{{{-x^5}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C\]
D
\[{{ {x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C\]
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Official solution
\[\int {{{2{x^{12}} + 5{x^9}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx\] Dividing by
x
15
{x^{15}}
x
15
in numerator and denominator \[\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over 5}} \right)}^3}}}} \] Substitute \[1 + {1 \over {{x^2}}} + {1 \over {{x^5}}} = t\] \[ \Rightarrow \left( {{{ - 2} \over {{x^3}}} - {5 \over {{x^6}}}} \right)dx = dt\] \[ \Rightarrow \left( {{2 \over {{x^3}}} + {5 \over {{x^6}}}} \right)dx = - dt\] This gives, \[\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over {{x^5}}}} \right)}^3}}}} \] \[ = \int {{{ - dt} \over {{t^3}}}} = {1 \over {2{t^2}}} + C\] \[ = {1 \over {2{{\left( {1 + {1 \over {{x^2}}} + {1 \over {{x^5}}}} \right)}^2}}} + C\] \[ = {{{x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C\]
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