Question:
The integral
\[
\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}} \, dx
\]
is equal to:
The integral
\[
\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}} \, dx
\]
is equal to:
Show Hint
When an integral contains powers of \((x-1)\) and \((x+2)\), try differentiating \(\left(\frac{x-1}{x+2}\right)^n\).
Updated On: May 20, 2026
- \(\displaystyle \frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{1}{4}}+C\)
- \(\displaystyle \frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{5}{4}}+C\)
- \(\displaystyle \frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C\)
- \(\displaystyle \frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{5}{4}}+C\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
For integrals containing powers of \((x-1)\) and \((x+2)\), a useful substitution is: \(\displaystyle u=\frac{x-1}{x+2}\) This is useful because the derivative of this fraction produces a factor involving \((x+2)^2\).
Step 1: Rewrite the integral.
\(\displaystyle I=\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}}\,dx\) This can be written as: \(\displaystyle I=\int (x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}\,dx\)
Step 2: Choose a suitable expression.
Let us check the derivative of: \(\displaystyle \left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}\)
Step 3: Differentiate the inner fraction.
Let, \(\displaystyle u=\frac{x-1}{x+2}\) Then, \(\displaystyle \frac{du}{dx}=\frac{(x+2)(1)-(x-1)(1)}{(x+2)^2}\) \(\displaystyle \frac{du}{dx}=\frac{x+2-x+1}{(x+2)^2}\) \(\displaystyle \frac{du}{dx}=\frac{3}{(x+2)^2}\)
Step 4: Differentiate the complete expression.
\(\displaystyle \frac{d}{dx}\left[\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}\right]\) \(\displaystyle =\frac{1}{4}\left(\frac{x-1}{x+2}\right)^{-\frac{3}{4}}\cdot \frac{3}{(x+2)^2}\) \(\displaystyle =\frac{3}{4}(x-1)^{-\frac{3}{4}}(x+2)^{\frac{3}{4}}\cdot (x+2)^{-2}\) \(\displaystyle =\frac{3}{4}(x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}\)
Step 5: Compare with the given integrand.
The given integrand is: \(\displaystyle (x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}\) We obtained: \(\displaystyle \frac{d}{dx}\left[\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}\right]=\frac{3}{4}(x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}\) Therefore, \(\displaystyle (x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}=\frac{4}{3}\frac{d}{dx}\left[\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}\right]\)
Step 6: Write the final integral.
\(\displaystyle I=\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C\) Hence, \(\displaystyle \boxed{\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C}\)
For integrals containing powers of \((x-1)\) and \((x+2)\), a useful substitution is: \(\displaystyle u=\frac{x-1}{x+2}\) This is useful because the derivative of this fraction produces a factor involving \((x+2)^2\).
Step 1: Rewrite the integral.
\(\displaystyle I=\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}}\,dx\) This can be written as: \(\displaystyle I=\int (x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}\,dx\)
Step 2: Choose a suitable expression.
Let us check the derivative of: \(\displaystyle \left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}\)
Step 3: Differentiate the inner fraction.
Let, \(\displaystyle u=\frac{x-1}{x+2}\) Then, \(\displaystyle \frac{du}{dx}=\frac{(x+2)(1)-(x-1)(1)}{(x+2)^2}\) \(\displaystyle \frac{du}{dx}=\frac{x+2-x+1}{(x+2)^2}\) \(\displaystyle \frac{du}{dx}=\frac{3}{(x+2)^2}\)
Step 4: Differentiate the complete expression.
\(\displaystyle \frac{d}{dx}\left[\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}\right]\) \(\displaystyle =\frac{1}{4}\left(\frac{x-1}{x+2}\right)^{-\frac{3}{4}}\cdot \frac{3}{(x+2)^2}\) \(\displaystyle =\frac{3}{4}(x-1)^{-\frac{3}{4}}(x+2)^{\frac{3}{4}}\cdot (x+2)^{-2}\) \(\displaystyle =\frac{3}{4}(x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}\)
Step 5: Compare with the given integrand.
The given integrand is: \(\displaystyle (x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}\) We obtained: \(\displaystyle \frac{d}{dx}\left[\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}\right]=\frac{3}{4}(x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}\) Therefore, \(\displaystyle (x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}=\frac{4}{3}\frac{d}{dx}\left[\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}\right]\)
Step 6: Write the final integral.
\(\displaystyle I=\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C\) Hence, \(\displaystyle \boxed{\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C}\)
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