Question

$\int \frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{3x}}dx$ is equal to

• A. $6\{\frac{x}{6}+\frac{x^{6/5}}{5}+\frac{x^{1/2}}{2}+\frac{x^{1/3}}{3}+\log (x^{1/6}-1)\}+c$
• B. $6\{\frac{x}{6}+\frac{x^{6/5}}{5}+\frac{x^{1/2}}{3}+\frac{x^{1/3}}{2}+\log (x^{1/6}-1)\}+c$
• C. $6\{\frac{x}{6}+\frac{x^{6/5}}{5}+\frac{x^{1/2}}{2}+\frac{x^{1/3}}{3}+x^{1/6}+\log (x^{1/6}-1)\}+c$
• D. None of the above

Answer 1

The correct option is D None of the above

Let $I=\int \frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{3x}}dx$

or $\int \frac{x^{1/2}}{x^{1/2}-x^{1/3}}dx$

Let $x=t^6$

$\Rightarrow dx=6t^{5}dt$

$\therefore I=\int \frac{t^3}{t^3-t^2}6t^{5}dt$

$=6\int \frac{t^6}{t-1}dt=6\int \frac{t^6-1+1}{t-1}dt$

After dividing we get,

$I=6\int (t^5+t^4+t^3+t^2+t+1+\frac{1}{t-1})dt$

$=6[\frac{t^6}{6}+\frac{t^5}{5}+\frac{t^4}{4}+\frac{t^3}{3}+\frac{t^2}{2}+t+\log (t-1)]+c$

$=6[\frac{x}{6}+\frac{x^{5/6}}{5}+\frac{x^{2/3}}{4}+\frac{x^{1/2}}{3}+\frac{x^{1/3}}{2}+x^{1/6}+\log (x^{1/6}-1)]+c[\because x=t^6\Rightarrow t=x^{1/6}]$