Question

The value of $\int (\frac{1}{\sqrt[3]{3x}+\sqrt[4]{4x}}+\frac{\log (1+\sqrt[6]{6x})}{\sqrt[3]{3x}+\sqrt{x}})dx$, is equal to $\dots \dots$ where $t=x^{\frac{1}{12}}$ and $\theta =\log (1+x^{\frac{1}{6}})$

• A. $12\{\sum _{r=1}^8{}^8{C}_r{(-1)}^r\frac{t^r}{r}+\log (1+t)\}+6\{(\frac{\theta }{3}-\frac{1}{9})e^{3\theta }-\frac{3}{2}(\theta -\frac{1}{2})e^{2\theta }+3(\theta -1)e^{\theta }-\frac{{\theta }^2}{2}\}+C$
• B. $6\{\sum _{r=1}^8{}^8{C}_r{(-1)}^r\frac{t^r}{r}+\log (1+t)\}+12\{(\frac{\theta }{3}-\frac{1}{9})e^{3\theta }-\frac{3}{2}(\theta -\frac{1}{2})e^{2\theta }+3(\theta -1)e^{\theta }-\frac{{\theta }^2}{2}\}+C$
• C. $\{\sum _{r=1}^8{}^8{C}_r{(-1)}^r\frac{t^r}{r}+\log (1+t)\}+6\{(\frac{\theta }{3}-\frac{1}{9})e^{3\theta }-\frac{3}{2}(\theta -\frac{1}{2})e^{2\theta }+3(\theta -1)e^{\theta }-\frac{{\theta }^2}{2}\}+C$
• D. none of the above

Answer 1

The correct option is A $12\{\sum _{r=1}^8{}^8{C}_r{(-1)}^r\frac{t^r}{r}+\log (1+t)\}+6\{(\frac{\theta }{3}-\frac{1}{9})e^{3\theta }-\frac{3}{2}(\theta -\frac{1}{2})e^{2\theta }+3(\theta -1)e^{\theta }-\frac{{\theta }^2}{2}\}+C$

This integration can be split in two parts
$\int \frac{1}{\sqrt[3]{3x}+\sqrt[4]{4x}}$
$t^{12}=x$
$12\int \frac{t^{11}dt}{t^4+t^3}$
Take $t+1=z$
$t^8={(z-1)}^8$
Expand Using Binomial Theorem ,integrate to get
$12\{\sum _{r=1}^8{}^8{C}_r{(-1)}^r\frac{t^r}{r}+\log (1+t)\}$ (1)
Another part can be written as
$\frac{12.t^{11}\log (1+t^2)dt}{t^4(t^2+1)}$
$\log (1+t^2)=u,\frac{2tdt}{1+t^2}=du$
$\int 6u{t}^{6}du=\int (u{e}^{3u}-u-3u{e}^{2u}+3u{e}^u)du$
$=6(\frac{u{e}^{3u}}{3}-\frac{1}{9}{e}^{3u}-\frac{u^2}{2}+3u{e}^u+3e^u-\frac{3u{e}^{2u}}{2}+\frac{3}{4}{e}^{2u}$)
$=6(\frac{u{e}^{3u}}{3}-\frac{1}{9}{e}^{3u}-\frac{u^2}{2}+3u{e}^u+3e^u-\frac{3u{e}^{2u}}{2}+\frac{3}{4}{e}^{2u})$
$=6\{(\frac{\theta }{3}-\frac{1}{9})e^{3\theta }-\frac{3}{2}(\theta -\frac{1}{2})e^{2\theta }+3(\theta -1)e^{\theta }-\frac{{\theta }^2}{2}\}+C$