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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
$\smallint \frac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}dx = $
  • A
    $\frac{{{x^5}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + c$
  • B
    $\frac{{ - {x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + c$
  • C
    $\frac{{ - {x^5}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + c$
  • ✓
    $\frac{{{x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + c$
✓

Answer

Correct option: D.
$\frac{{{x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + c$
d
$\int {\frac{{2{x^{12}} + 5{x^9}}}{{{{\left[ {{x^5}\left( {1 + \frac{1}{{{x^2}}} + \frac{1}{{{x^5}}}} \right)} \right]}^3}}}}  = \int {\frac{{2{x^{12}} + 5{x^9}}}{{{x^{15}}{{\left( {1 + \frac{1}{{{x^2}}} + \frac{1}{{{x^5}}}} \right)}^3}}}} dx$

Dividing numerator and denominator by $x^{15}$ we get,

$ = \int {\frac{{\frac{2}{{{x^3}}} + \frac{5}{{{x^6}}}}}{{{{\left( {1 + \frac{1}{{{x^2}}} + \frac{1}{{{x^5}}}} \right)}^3}}}} dx$

Put $\left(1+\frac{1}{x^{2}}+\frac{1}{x^{5}}\right)=t$

$=\int \frac{-d t}{t^{3}}$

$=\frac{-t^{-3+1}}{-3+1}+C=\frac{1}{2} \times \frac{1}{t^{2}}+C$

$=\frac{1}{2} \frac{1}{\left(1+\frac{1}{x^{2}}+\frac{1}{x^{5}}\right)^{2}}+C$

$=\frac{1}{2} \frac{x^{10}}{\left(x^{5}+x^{3}+1\right)^{2}}+C$

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