The integral ∫ 1 √[4](x-1)^3(x+2)^5dx is equal to : (where C is a constant of integration)

$$I=\int \frac{dx}{\sqrt[4]{(x-1{)}^{3}(x+2{)}^{5}}}$$I=∫4(x−1)3(x+2)5​dx​

$$I=\int \frac{dx}{(x-1{)}^{\frac{3}{4}}(x+2{)}^{\frac{5}{4}}}$$I=∫(x−1)43​(x+2)45​dx​

$$I=\int \frac{dx}{(x-1{)}^{2}{(\frac{x+2}{x-1})}^{\frac{5}{4}}}$$I=∫(x−1)2(x−1x+2​)45​dx​

Let $\frac{x+2}{x-1}=t$x−1x+2​=t

Differentiate on both sides

$$\frac{(x-1)-(x+2)}{{(x-1)}^{2}}.dx=dt$$(x−1)2(x−1)−(x+2)​.dx=dt

$$\Rightarrow \frac{dx}{{(x-1)}^{2}}=\frac{-dt}{3}$$⇒(x−1)2dx​=3−dt​

$$I=\frac{-1}{3}\int \frac{dt}{{(t)}^{\frac{5}{4}}}$$I=3−1​∫(t)45​dt​

$$=\frac{-1}{3}\frac{{t}^{\frac{-5}{4}+1}}{1-\frac{5}{4}}+C$$=3−1​1−45​t4−5​+1​+C

$$=\frac{4}{3}{t}^{\frac{-1}{4}}+C$$=34​t4−1​+C

$$=\frac{4}{3}{(\frac{1}{t})}^{\frac{1}{4}}+C$$=34​(t1​)41​+C

$$=\frac{4}{3}{(\frac{x-1}{x+2})}^{\frac{1}{4}}+C$$=34​(x+2x−1​)41​+C