First slide

Methods of integration

Question

$\int \frac{1}{{[(x - 1)^{3} (x + 2)^{5}]}^{1 / 4}} d x is equal to$

Moderate

A

$\frac{4}{3} {(\frac{x - 1}{x + 2})}^{1 / 4} + C$
B

$\frac{4}{3} {(\frac{x + 2}{x - 1})}^{1 / 4} + C$
C

$\frac{1}{3} {(\frac{x - 1}{x + 2})}^{1 / 4} + C$
D

$\frac{1}{3} {(\frac{x + 2}{x - 1})}^{1 / 4} + C$

Solution

$\begin{array}{l} l = \int \frac{1}{{[(x - 1)^{3} (x + 2)^{5}]}^{1 / 4}} d x \\ = \int \frac{1}{{(\frac{x - 1}{x + 2})}^{3 / 4} (x + 2)^{2}} d x \\ Let \frac{x - 1}{x + 2} = t \Rightarrow \frac{3 d x}{(x + 2)^{2}} = d t \\ \Rightarrow l = \frac{1}{3} \int \frac{1}{t^{3 / 4}} d t \\ = \frac{1}{3} (\frac{t^{1 / 4}}{1 / 4}) + C = \frac{4}{3} t^{1 / 4} + C = \frac{4}{3} {(\frac{x - 1}{x + 2})}^{1 / 4} + C \end{array}$

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