First slide

Methods of integration

Question

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

Moderate

A

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

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

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

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

Solution

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

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