The value of \int \frac{1}{x{1}} \dx, is [(x-1)^3(x+2)^5] ^{\frac{1}{4}}

Correct Answer is : 43(x−1x+2)14+C

\frac{4}{3} (\frac{x+1}{x-2}) ^{\frac{1}{4}}+C

\frac{3}{4} (\frac{x-1}{x+2}) ^{\frac{1}{4}}+C

\frac{4}{3} (\frac{x-1}{x+2}) ^{\frac{1}{4}}+C

\frac{1}{3} (\frac{2 x-1}{4 x-3}) ^{\frac{1}{4}}+C

BITSAT 2022 Paper

Section: Mathematics

Question : 105 of 130

Marks: +1, -0

The value of ∫‌

‌dx, is [(x−1)3(x+2)5]‌

Go to Question:

Prev Question

Next Question