Question

$\int \frac{dx}{x(1+\sqrt[3]{3x}{)}^2}$ is equal to

• A. $3(\log \frac{x^{1/3}}{1+x^{1/3}}+\frac{1}{1+\sqrt[3]{3x}})+c$
• B. $3(\log \frac{1+\sqrt[3]{3x}}{\sqrt[3]{3x}}+\frac{1}{1+\sqrt[3]{3x}})+c$
• C. $3(\log \frac{x^{1/3}}{1+x^{1/3}}-\frac{1}{1+\sqrt[3]{3x}})+c$
• D. $3(\log \frac{1+\sqrt[3]{3x}}{\sqrt[3]{3x}}-\frac{1}{1+\sqrt[3]{3x}})+c$

Answer 1

Answer: (A)

$3(\log \frac{x^{1/3}}{1+x^{1/3}}+\frac{1}{1+\sqrt[3]{3x}})+c$

$I=\int \frac{dx}{x(1+\sqrt[3]{3x}{)}^2}$
Put $x^{1/3}=t$
$\frac{1}{3}{x}^{-2/3}dx=dt$
$\Rightarrow dx=3t^{2}dt$
So, $I=3\int \frac{t^{2}dt}{t^3(1+t{)}^2}$
$I=3\int \frac{dt}{t(1+t{)}^2}$
Resolving $\frac{1}{t(1+t{)}^2}$ into partial fractions
$\frac{1}{t(1+t{)}^2}=\frac{A}{t}+\frac{B}{(1+t)}+\frac{C}{(1+t{)}^2}$ .....(1)
$\Rightarrow 1=A(1+t{)}^2+B(1+t)+Ct$
$\Rightarrow A=1,B=-1,C=-1$
Put these values in (1),
$\frac{1}{t(1+t{)}^2}=\frac{1}{t}-\frac{1}{(1+t)}-\frac{1}{(1+t{)}^2}$
$\int \frac{1}{t(1+t{)}^2}dt=\int \frac{1}{t}dt-\int \frac{1}{(1+t)}dt-\int \frac{1}{(1+t{)}^2}dt$
$\int \frac{1}{t(1+t{)}^2}dt=\log |t|-\log |1+t|+\frac{1}{1+t}+C$
So, $I=3(\log \frac{x^{1/3}}{1+x^{1/3}}+\frac{1}{1+\sqrt[3]{3x}})+c$