Discussion Forum

If  $\int\frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{x}}dx=x+Ex^{5/6}+Dx^{2/3}+Cx^{1/2}+Bx^{1/3}+Ax^{1/6}+\log (\sqrt[6]{x}-1)^{6}+K,$ then A+B+C+D+E=

Answer:

Explanation:

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

 Put   $x=t^{6}\Rightarrow dx=6t^{5}dx$

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

$=6\int\frac{t^{8}}{t^{2}(t-1)}dt=6\int\frac{t^{6}}{t-1}dt$

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

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

 $=6\int\left[ t^{5}+t^{2}+t^{4}+t+t^{3}+1+\frac{1}{t-1}\right]dt$

$=6\int\left[ t^{5}+t^{4}+t^{3}+t^{2}+t+1+\frac{1}{t-1}\right]dt$

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

 $= x+ \frac{6}{5}x^{5/6}+\frac{3}{2}x^{2/3}+2x^{1/2}+3x^{1/3}+6x^{1/6}+\log (\sqrt[6]{x-1})^{6}+k$

 $\therefore$    $A=6,B=3,C=2,D=\frac{3}{2},E=\frac{6}{5}$

 $therefore$     $A+B+C+D+E=6+3+2+\frac{3}{2}+\frac{6}{5}$

$=11+\frac{3}{2}+\frac{6}{5}= \frac{110+15+12}{10}=\frac{137}{10}$