Question

The value of $\int \frac{dx}{{(x+1)}^{\frac{1}{3}}+{(x+1)}^{\frac{1}{2}}}$, is equal to

• A. $t^3-3t^2+t-6\log |1+t|+C$; where $t={(x+1)}^{\frac{1}{6}}$
• B. $2t^3-3t^2+6t-6\log |1+t|+C$; where $t={(x+1)}^{\frac{1}{6}}$
• C. $t^3-3t^2+6t-6\log |1+t|+C$; where $t={(x^2-1)}^{\frac{1}{6}}$
• D. none of the above

Answer 1

Answer: (B)

$2t^3-3t^2+6t-6\log |1+t|+C$; where $t={(x+1)}^{\frac{1}{6}}$

Let $I=\int \frac{1}{\sqrt[3]{3x+1}+\sqrt[2]{2x+1}}dx$
Putting $u=x+1\Rightarrow du=dx$

$I=\int \frac{1}{\sqrt[3]{3u}+\sqrt[2]{2u}}du$

Putting $\sqrt[6]{6u}=s\Rightarrow ds=\frac{1}{6u^{\frac{5}{6}}}du$

$I=6\int \frac{s^5}{s^3+s^2}ds=6\int \frac{s^3}{s+1}ds=6\int (s^2-s-\frac{1}{s+1}+1)ds$

$=2s^3-3s^2+6s-6\log (s+1)+c=2\sqrt{u}-3\sqrt[3]{3u}+6\sqrt[6]{6u}-6\log (\sqrt[6]{6u}+1)+c=2\sqrt{x+1}-3\sqrt[3]{3x+1}+6\sqrt[6]{6x+1}-6\log (\sqrt[6]{6x+1}+1)+c$

Hence $I=2t^3-3t^2+6t-6\log |1+t|+c$; where $t=\sqrt[6]{6x+1}$