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The integral $\int \frac{dx}{(x+1)^{\frac{3}{4}}(x-2)^{\frac{5}{4}}}$

is equal to:

Option 1)
$4\left ( \frac{x+1}{x-2} \right )^{\frac{1}{4}}+C$

Option 2)
$4\left ( \frac{x-2}{x+1} \right )^{\frac{1}{4}}+C$

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

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

Answers (1)

best_answer

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

- wherein

Since $\int f(x)dx=\int f(t)dt=\int f(\theta )d\theta$ all variables must be converted into single variable , $\left ( t\, or\ \theta \right )$

$I = \frac{dx}{\left ( x+1 \right )^\frac{3}{4} \left ( x-2 \right )^{\frac{5}{4}}}$

$I = \int \frac{dx}{\left ( x+1 \right )^\frac{3}{4} \left ( x-2 \right )^\frac{5}{4}} \frac{\left ( x-2 \right )^\frac{3}{4}}{\left ( x-2 \right )^{\frac{3}{4}}}$

$= \int \frac{dx}{\left ( \frac{x+1}{x-2} \right )^\frac{3}{4} \left ( x-2 \right )^2}$

Let $\frac{x+1}{x-2}=t$

$\frac{\left ( x-2 \right )-(x+1)}{\left ( x-2 \right )^2}dx$

$\Rightarrow \frac{dx}{\left ( x-2 \right )^2}=-\frac{dt}{3}$

Thus I= $\int \frac{-dt}{3 t^\frac{3}{4}}$

$=\frac{-1}{3} \times \frac{1}{\frac{1}{4}} t^\frac{1}{4} +c$

$=-\frac{4}{3}t^\frac{1}{4}+c$

$I=-\frac{4}{3}\left ( \frac{x+1}{x-2} \right )^\frac{1}{4}+c$

Option 1)

$4\left ( \frac{x+1}{x-2} \right )^{\frac{1}{4}}+C$

Incorrect Option

Option 2)

$4\left ( \frac{x-2}{x+1} \right )^{\frac{1}{4}}+C$

Incorrect Option

Option 3)

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

Correct Option

Option 4)

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

Incorrect Option

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