Question:
If m is a non-zero number and \(\displaystyle\int\frac{x^{5m-1}+2x^{4m-1}}{\left(x^{2m}+x^m+1\right)^3}\,dx=f(x)+c\), then \(f(x)\) is :
If m is a non-zero number and \(\displaystyle\int\frac{x^{5m-1}+2x^{4m-1}}{\left(x^{2m}+x^m+1\right)^3}\,dx=f(x)+c\), then \(f(x)\) is :
Show Hint
Divide numerator and denominator by \(x^{6m}\) and substitute \(t=x^{-2m}+x^{-m}+1\).
Updated On: Nov 11, 2026
- \(\frac{x^{5m}}{2m\left(x^{2m}+x^m+1\right)^2}\)
- \(\frac{x^{4m}}{2m\left(x^{2m}+x^m+1\right)^2}\)
- \(\frac{2m\left(x^{5m}+x^{4m}\right)}{\left(x^{2m}+x^m+1\right)^2}\)
- \(\frac{x^{5m}-x^{4m}}{2m\left(x^{2m}+x^m+1\right)^2}\)
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The Correct Option is B
Solution and Explanation
Divide numerator and denominator by \(x^{6m}\): integrand \(=\frac{x^{-m-1}+2x^{-2m-1}}{\left(x^{-2m}+x^{-m}+1\right)^3}\).
Let \(t=x^{-2m}+x^{-m}+1\). Then \(dt=-m\left(2x^{-2m-1}+x^{-m-1}\right)dx\).
So the integral \(=-\frac1m\int t^{-3}dt=\frac{1}{2m\,t^2}+c\).
\(\frac{1}{2mt^2}=\frac{x^{4m}}{2m\left(x^{2m}+x^m+1\right)^2}\) (since \(t=\frac{x^{2m}+x^m+1}{x^{2m}}\)).
Answer: (2) \(\frac{x^{4m}}{2m\left(x^{2m}+x^m+1\right)^2}\)
Let \(t=x^{-2m}+x^{-m}+1\). Then \(dt=-m\left(2x^{-2m-1}+x^{-m-1}\right)dx\).
So the integral \(=-\frac1m\int t^{-3}dt=\frac{1}{2m\,t^2}+c\).
\(\frac{1}{2mt^2}=\frac{x^{4m}}{2m\left(x^{2m}+x^m+1\right)^2}\) (since \(t=\frac{x^{2m}+x^m+1}{x^{2m}}\)).
Answer: (2) \(\frac{x^{4m}}{2m\left(x^{2m}+x^m+1\right)^2}\)
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