First slide

Methods of integration

Question

$\int \frac{p x^{p + 2 q - 1} - q x^{q - 1}}{x^{2 p + 2 q} + 2 x^{p + q} + 1} dx is equal to$

Difficult

A

$- \frac{x^{p}}{x^{p + q} + 1} + C$
B

$\frac{x^{q}}{x^{p + q} + 1} + C$
C

$- \frac{x^{q}}{x^{p + q} + 1} + C$
D

$\frac{x^{p}}{x^{p + q} + 1} + C$

Solution

$\begin{array}{l} \int \frac{p x^{p + 2 q - 1} - q x^{q - 1}}{{(x^{p + q} + 1)}^{2}} d x \\ (Dividing N^{r} and D^{r} by x^{2 q}) \\ = \int \frac{p x^{p - 1} - q x^{- q - 1}}{{(x^{p} + x^{- q})}^{2}} d x \\ t = x^{p} + x^{- q}, d t = (p x^{p - 1} - q x^{- q - 1}) d x \\ = \int \frac{d t}{t^{2}} = \frac{1}{t} + C \\ = - \frac{1}{x^{p} + x^{- q}} + C \\ = - \frac{x^{q}}{x^{p + q} + 1} + C \end{array}$

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