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MHT CET · Maths · Indefinite Integration

The value of \(\int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} \mathrm{~d} x\) is equal to (where \(C\) is arbitrary constant.)

  1. A \(\frac{x^5}{2\left(x^5+x^3+1\right)^2}+C\)
  2. B \(\frac{x^{10}}{2\left(x^5+x^3+1\right)^2}+C\)
  3. C \(\frac{-x^5}{\left(x^5+x^3+1\right)^2}+C\)
  4. D \(\frac{-x^{10}}{2\left(x^5+x^3+1\right)^2}+C\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} \mathrm{~d} x=\int \frac{2 x^{12}+5 x^9}{x^{15}\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^3} \mathrm{~d} x \\ & =\int \frac{\frac{2}{x^3}+\frac{5}{x^6}}{\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^3} \mathrm{~d} x \\ & =\int \frac{-\mathrm{d} t}{t^3}=\frac{1}{2 t^2}+C\left[\operatorname{let} 1+\frac{1}{x^2}+\frac{1}{x^5}=t\right] \\ & =\frac{1}{2\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^2}+C \\ & =\frac{x^{10}}{2\left(x^5+x^3+1\right)^2}+C\end{aligned}\)
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