JEE Mains · Maths · STD 12 - 7.1 indefinite integral
The integral \(\int \frac{1}{\sqrt[4]{(x-1)^{3}(x+2)^{5}}} d x\) is equal to : (where \(\mathrm{C}\) is a constant of integration)
- A \(\frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{1}{4}}+C\)
- B \(\frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{5}{4}}+C\)
- C \(\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C\)
- D \(\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{5}{4}}+C\)
Answer & Solution
Correct Answer
(C) \(\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C\)
Step-by-step Solution
Detailed explanation
\(\int \frac{d x}{(x-1)^{3 / 4}(x+2)^{5 / 4}}\) \(=\int \frac{d x}{\left(\frac{x+2}{x-1}\right)^{5 / 4} \cdot(x-1)^{2}}\) put \(\frac{x+2}{x-1}=t\) \(=-\frac{1}{3} \int \frac{d t}{t^{5 / 4}}\) \(=\frac{4}{3} \cdot \frac{1}{t^{1 / 4}}+C\)…
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