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)
D Divya Saini

 

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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