Question

The integral $I = \int {\dfrac{{dx}}{{{{\left( {x + 1} \right)}^{3/4}} * {{\left( {x - 2} \right)}^{5/4}}}}} $ is equal to:
A.$4{\left( {\dfrac{{x + 1}}{{x - 2}}} \right)^{1/4}} + C$
B.$4{\left( {\dfrac{{x - 2}}{{x + 2}}} \right)^{1/4}} + C$
C.$\dfrac{{ - 4}}{3}{\left( {\dfrac{{x + 1}}{{x - 2}}} \right)^{1/4}} + C$
D.$\dfrac{{ - 4}}{3}{\left( {\dfrac{{x - 2}}{{x + 2}}} \right)^{1/4}} + C$

Answer 1

Hint: In mathematics, Integral equations are which an unknown function appears under an integral sign. Integration is the algebraic method to find the integral for a function at any point of graph. Hence integral is also called the anti-derivative because integration is a reverse process of differentiation. In the given question, first of all we will break the given term $\left( {x - 2} \right)$ and assume it as$t$, then put this value in the given equation and differentiate it with respect to $t$ after that integrating the obtained result we will get the answer.

Complete step-by-step answer:
Given that:
$I = \int {\dfrac{{dx}}{{{{\left( {x + 1} \right)}^{3/4}} *{{\left( {x - 2} \right)}^{5/4}}}}} $
We can write the above equation as:
$\int {\dfrac{1}{{{{\left( {x - 2} \right)}^{2/4}} * {{\left( {x + 1} \right)}^{3/4}}}}} \times \dfrac{{dx}}{{{{\left( {x - 2} \right)}^{3/4}}}}$
Here we break the term $\left( {x - 2} \right)$
Let
${\left( {x - 2} \right)^{1/4}} = t..............\left( 1 \right)$
$
   \Rightarrow \left( {x - 2} \right) = {t^{4}} \\
   \Rightarrow x = {t^4} + 2 \\
 $
From equation one:
$
  \dfrac{1}{4}{\left( {x - 2} \right)^{ - 3/4}}dx = dt \\
   \Rightarrow \dfrac{{dx}}{{{{\left( {x - 2} \right)}^{ 3/4}}}} = 4dt \\
 $
Integrating the above equation
We get:
$\int {\dfrac{{4dt}}{{{t^2}{{\left( {{t^4} + 3} \right)}^{^{3/4}}}}}} $
Substitute $\dfrac{{{{\left( {{t^4} + 3} \right)}^{1/4}}}}{t} = u$ …………….. (2)
Now differentiate on both sides of the above equation:
\[
  \left[ {\dfrac{{t\dfrac{1}{4}{{\left( {{t^4} + 3} \right)}^{ - 3/4}}4{t^3} - {{\left( {{t^4} + 3} \right)}^{1/4}}}}{{{t^2}}}} \right]dt = du \\
   \Rightarrow \left[ {\dfrac{{{t^4}}}{{{{\left( {{t^4} + 3} \right)}^{3/4}}}} - {{\left( {{t^4} + 3} \right)}^{1/4}}} \right]/{t^2}dt = du \\
   \Rightarrow \dfrac{{{t^4}\left( {{t^4} + 3} \right)}}{{{t^2}{{\left( {{t^4} + 3} \right)}^{3/4}}}}dt = du \\
   \Rightarrow \dfrac{{ - 3dt}}{{{t^2}{{\left( {{t^4} + 3} \right)}^{3/4}}}} = du \\
 \]
Substitute the value of u from equation 2
$
  \dfrac{{ - 4}}{3}\int {du} \\
   \Rightarrow \dfrac{{ - 4}}{3}u + C \\
 $
Put the value of u from equation 2
We get:
$\dfrac{{ - 4}}{3}\left[ {\dfrac{{{{\left( {{t^4} + 3} \right)}^{3/4}}}}{t}} \right] + C$
Now we will put the value of t from equation 1
We get
\[
   \Rightarrow \dfrac{{ - 4}}{3}\left[ {\dfrac{{{{\left( {x - 2} \right)}^{1/4}} + 3}}{{{{\left( {x - 2} \right)}^{1/4}}}}} \right] + C \\
   \Rightarrow \dfrac{{ - 4}}{3}\left[ {\dfrac{{{{\left( {x - 2 + 3} \right)}^{1/4}}}}{{{{\left( {x - 2} \right)}^{1/4}}}}} \right] + C \\
   \Rightarrow \dfrac{{ - 4}}{3}\left[ {\dfrac{{{{\left( {x + 1} \right)}^{1/4}}}}{{{{\left( {x - 2} \right)}^{1/4}}}}} \right] + C \\
   \Rightarrow \dfrac{{ - 4}}{3}\left[ {{{\left( {\dfrac{{x + 1}}{{x - 2}}} \right)}^{1/4}}} \right] + C \\
 \]
Hence the correct answer is option C

Note: The given problem we have to break the equation, without this we can’t solve the equation. After that assume the common term as $t$ and integrate it with respect to$t$. Consider the outcome as $u$ and differentiate it. We have to add a constant integration symbol c and put the value of $t$ in it. Thus we get the answer.