\( \int\displaystyle\frac{1}{(x-1)^{3/4}(x+2)^{5/4}}\,dx \) is equal to :
[where C is an arbitrary constant]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{1/4}+C \)
The powers add up to \( \frac{3}{4}+\frac{5}{4}=2 \), so the integrand can be written in terms of \( \frac{x-1}{x+2} \):
\( \displaystyle \frac{1}{(x-1)^{3/4}(x+2)^{5/4}}=\left(\frac{x-1}{x+2}\right)^{-3/4}\cdot\frac{1}{(x+2)^2} \)
Let
\( \displaystyle t=\frac{x-1}{x+2}=1-\frac{3}{x+2} \)
\( \displaystyle dt=\frac{3}{(x+2)^2}\,dx \)
Then
\( \displaystyle \int\left(\frac{x-1}{x+2}\right)^{-3/4}\frac{dx}{(x+2)^2}=\frac{1}{3}\int t^{-3/4}\,dt \)
\( \displaystyle =\frac{1}{3}\cdot\frac{t^{1/4}}{1/4}+C \)
\( \displaystyle =\frac{4}{3}\,t^{1/4}+C \)
\( \displaystyle =\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{1/4}+C \)
Tip: For \( \int\frac{dx}{(x-a)^m(x-b)^n} \) with m + n = 2, the substitution \( t=\frac{x-a}{x-b} \) always works, because \( dt \) is a constant times \( \frac{dx}{(x-b)^2} \).