∫x5m−1+2x4m−1x2m+xm+13dx is equal to where m∈R−{0}

I=∫x5m−1+2x4m−1dxx6m1+x−m+x−2m3=∫x−(m+1)+2⋅x−(2m+1)1+x−m+x−2m3dx Put 1+x−m+x−2m=t-mx-m-1-2mx-2m-1dx=dt∴ I=−1m∫dtt3=-1mt-3+1-3+1=12mt2+c=x4m2mx2m+xm+12+c

∫x5m−1+2x4m−1x2m+xm+13dx is equal to where m∈R−{0}

I=∫x5m−1+2x4m−1dxx6m1+x−m+x−2m3=∫x−(m+1)+2⋅x−(2m+1)1+x−m+x−2m3dx Put 1+x−m+x−2m=t-mx-m-1-2mx-2m-1dx=dt∴ I=−1m∫dtt3=-1mt-3+1-3+1=12mt2+c=x4m2mx2m+xm+12+c

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$\int \frac{x^{5 m - 1} + 2 x^{4 m - 1}}{{(x^{2 m} + x^{m} + 1)}^{3}} dx is equal to where m \in R - {0}$

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$\frac{x^{6 m} + x^{2 m}}{2 m {(x^{2 m} + x^{m} + 1)}^{2}} + c$

$\frac{x^{4 m}}{2 m {(x^{2 m} + x^{m} + 1)}^{2}} + c$

$\frac{2 m x^{4 m}}{{(x^{2 m} + x^{m} + 1)}^{2}} + c$

$\frac{m x^{5 m}}{2 {(x^{2 m} + x^{m} + 1)}^{2}} + c$

answer is B.

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Detailed Solution

$\begin{array}{l} I = \int \frac{(x^{5 m - 1} + 2 x^{4 m - 1}) d x}{x^{6 m} {(1 + x^{- m} + x^{- 2 m})}^{3}} = \int \frac{x^{- (m + 1)} + 2 \cdot x^{- (2 m + 1)}}{{(1 + x^{- m} + x^{- 2 m})}^{3}} d x \\ Put (1 + x^{- m} + x^{- 2 m}) = t \\ (- m x^{- m - 1} - 2 m x^{- 2 m - 1}) d x = d t \\ ∴ I = \frac{- 1}{m} \int \frac{d t}{t^{3}} = \frac{- 1}{m} \frac{t^{- 3 + 1}}{- 3 + 1} = \frac{1}{2 m t^{2}} + c = \frac{x^{4 m}}{2 m {(x^{2 m} + x^{m} + 1)}^{2}} + c \end{array}$