Question

The integral $\displaystyle\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}}\,dx$ is equal to: (where $C$ is a constant of integration)}

• A. $\dfrac{3}{4}\left(\dfrac{x+2}{x-1}\right)^{1/4} + C$
• B. $\dfrac{3}{4}\left(\dfrac{x+2}{x-1}\right)^{5/4} + C$
• C. $\dfrac{4}{3}\left(\dfrac{x-1}{x+2}\right)^{1/4} + C$
• D. $\dfrac{4}{3}\left(\dfrac{x-1}{x+2}\right)^{5/4} + C$

Hint:

For integrals with fractional powers of two linear factors, try the substitution $t = \dfrac{ax+b}{cx+d}$, which turns the integral into a simple power function.

Answer 1

The Correct Option is C

Step 1: Concept
Rewrite the denominator: $(x-1)^{3/4}(x+2)^{5/4} = (x+2)^2\!\left(\dfrac{x-1}{x+2}\right)^{3/4}$.

Step 2: Meaning
Let $t = \dfrac{x-1}{x+2}$. Then $dt = \dfrac{(x+2)-(x-1)}{(x+2)^2}dx = \dfrac{3}{(x+2)^2}dx$, so $\dfrac{dx}{(x+2)^2} = \dfrac{dt}{3}$.

Step 3: Analysis
The integral becomes: \[\int \frac{1}{(x+2)^2\cdot t^{3/4}}\,dx = \frac{1}{3}\int t^{-3/4}\,dt = \frac{1}{3}\cdot\frac{t^{1/4}}{1/4} + C = \frac{4}{3}t^{1/4} + C.\]

Step 4: Conclusion
Substituting back: $I = \dfrac{4}{3}\!\left(\dfrac{x-1}{x+2}\right)^{1/4} + C$. Final Answer: (C)