The integral \(\int \frac{1}{\sqrt[4]{(x-1)^3(x + 2)^5}}\)dx is equal to : (where C is a constant of integration)
(1) \(\frac{3}{4}\)\(\Big(\frac{x+2}{x-1}\Big)^{\frac{1}{4}}\) + C
(2) \(\frac{3}{4}\)\(\Big(\frac{x+2}{x-1}\Big)^{\frac{5}{4}}\) + C
(3) \(\frac{4}{3}\)\(\Big(\frac{x-1}{x+2}\Big)^{\frac{1}{4}}\) + C
(4) \(\frac{4}{3}\)\(\Big(\frac{x-1}{x+2}\Big)^{\frac{5}{4}}\) + C
Correct answer: (3) \(\frac{4}{3}\)\(\Big(\frac{x-1}{x+2}\Big)^{\frac{1}{4}}\) + C
Explanation:
\(\int \frac{dx}{(x-1)^{\frac{3}{4}}(x + 2)^{\frac{5}{4}}}\)
= \(\int \frac{dx}{(\frac{x+2}{x-1})^{\frac{5}{4}}.(x-1)^2}\)
put \(\frac{x+2}{x-1}\) = t
= -\(\frac{1}{3}\)\(\int \frac{dt}{t^{\frac{5}{4}}}\)
= \(\frac{4}{3}\).\(\frac{1}{t^{\frac{1}{4}}}\) + C
= \(\frac{4}{3}\)\(\Big(\frac{x-1}{x+2}\Big)^{\frac{1}{4}}\) + C