Question

lf $\int \frac{dx}{\sqrt[4]{4(x-1{)}^3(x+2{)}^5}}=A\sqrt[4]{4\frac{x-1}{x+2}}+c$, then $A$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $\frac{4}{3}$

Answer 1

Answer: (D)

$\frac{4}{3}$

Given, $\int \frac{dx}{\sqrt[4]{4(x-1{)}^3(x+2{)}^5}}=A\sqrt[4]{4\frac{x-1}{x+2}}+c$ ....(1)
Consider, $I=\int \frac{dx}{\sqrt[4]{4(x-1{)}^3(x+2{)}^5}}$
$=\int \frac{dx}{{(x-1)}^{3/4}{(x+2)}^{5/4}}$
$=\int \frac{dx}{{\left(\frac{x-1}{x+2}\right)}^{3/4}{(x+2)}^2}$
Put $\frac{x-1}{x+2}=t$
$\frac{3}{(x+2{)}^2}dx=dt$
So, $I=\frac{1}{3}\int \frac{dt}{t^{3/4}}$
$I=\frac{4}{3}{t}^{1/4}+c$
$I=\frac{4}{3}{\left(\frac{x-1}{x+2}\right)}^{1/4}+c$
On comparing with (1), we get
$A=\frac{4}{3}$