MHT CET · Maths · Indefinite Integration
The integral \(\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^4+3 x^2+1\right)^4} d x\) is equal to (where \(C\) is a constant of integration.)
- A \(\frac{x^{12}}{\left(2 x^4+3 x^2+1\right)^3}+C\)
- B \(\frac{x^4}{\left(2 x^4+3 x^2+1\right)^3}+C\)
- C \(\frac{x^4}{6\left(2 x^4+3 x^2+1\right)^3}+C\)
- D \(\frac{x^{12}}{6\left(2 x^4+3 x^2+1\right)^3}+C\)
Answer & Solution
Correct Answer
(D) \(\frac{x^{12}}{6\left(2 x^4+3 x^2+1\right)^3}+C\)
Step-by-step Solution
Detailed explanation
\(\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^4+3 x^2+1\right)^4} d x=\int \frac{3 x^{13}+2 x^{11}}{x^{16}\left(2+\frac{3}{x^2}+\frac{1}{x^4}\right)^4} d x\)
\(=-\frac{1}{2} \int \frac{-6 x^{-3}-4 x^{-5}}{\left(2+\frac{3}{x^2}+\frac{1}{x^4}\right)^4} d x=\frac{-1}{2} \times \frac{-1}{3}(2+\frac{3}{x^2}+\) \(\frac{1}{x^4})^{-3}+C\)
\(=\frac{1}{6}\left(\frac{2 x^4+3 x^2+1}{x^4}\right)^{-3}+C=\frac{x^{12}}{6\left(2 x^4+3 x^2+1\right)^3}+C\)
\(=-\frac{1}{2} \int \frac{-6 x^{-3}-4 x^{-5}}{\left(2+\frac{3}{x^2}+\frac{1}{x^4}\right)^4} d x=\frac{-1}{2} \times \frac{-1}{3}(2+\frac{3}{x^2}+\) \(\frac{1}{x^4})^{-3}+C\)
\(=\frac{1}{6}\left(\frac{2 x^4+3 x^2+1}{x^4}\right)^{-3}+C=\frac{x^{12}}{6\left(2 x^4+3 x^2+1\right)^3}+C\)
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