Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
If $\int \frac{dx}{\sqrt[3]{(x - 1)^3(x + 2)^2}} = k\sqrt[3]{\frac{x - 1}{x + 2}} + c$, then $3k$ is equal to
Step-by-Step Solution
Key Concept: The substitution $t = \frac{x-1}{x+2}$ converts the product of powers into a single power integral.
For $\int \frac{dx}{(x-1)^{3/4}(x+2)^{5/4}}$, use substitution $t = \frac{x-1}{x+2}$ to get $\frac{dt}{t^{3/4}} = \frac{4}{3}\frac{dx}{(x+2)^2}$. This simplifies to $I = \frac{1}{3}\int t^{-3/4}dt = \frac{1}{3} \cdot 4t^{1/4} = \frac{4}{3}\left(\frac{x-1}{x+2}\right)^{1/4} + c$.
Correct Answer: 1,2,3