First slide

Methods of integration

Question

$\int \frac{dx}{(x + 3)^{15 / 16} (x - 4)^{17 / 16}}$ is equal to

Moderate

A

${(\frac{x + 3}{x - 4})}^{\frac{1}{16}} + C$
B

$\frac{- 16}{7} {(\frac{x + 3}{x - 4})}^{\frac{1}{16}} + C$
C

$\frac{16}{5} {(\frac{x - 4}{x + 3})}^{\frac{1}{16}} + C$
D

None of these

Solution

$I = \int \frac{dx}{(x + 3)^{\frac{15}{16}} (x - 4)^{\frac{17}{16}}} = \int \frac{dx}{{(\frac{x + 3}{x - 4})}^{\frac{15}{16}} (x - 4)^{2}} Put \frac{x + 3}{x - 4} = t \Rightarrow [\frac{(x - 4) - (x + 3)}{(x - 4)^{2}}] dx = dt \Rightarrow \frac{dx}{(x - 4)^{2}} = \frac{dt}{- 7}$
So,
$\begin{array}{l} I = \frac{- 1}{7} \int \frac{dt}{t^{\frac{15}{16}}} = \frac{- 1}{7} \int t^{\frac{- 15}{16}} dt \\ = \frac{- 16}{7} t^{\frac{1}{16}} + C \\ = \frac{- 16}{7} {(\frac{x + 3}{x - 4})}^{1 / 16} + C \end{array}$

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