Question

If $\int \frac{dx}{\sqrt{x}+\sqrt[3]{3x}}=a\sqrt{x}+b\left(\sqrt[3]{3x}\right)+c\left(\sqrt[6]{6x}\right)+d\ln (\sqrt[6]{6x}+1)+e$, $e$ being arbitrary constant then. Find the value of $20a+b+c+d.$

• A. $37$
• B. $35$
• C. $43$
• D. $47$

Answer 1

The correct option is A $37$

$\int \frac{dx}{\sqrt{x}+\sqrt[3]{3x}}=a\sqrt{x}+b\left(\sqrt[3]{3x}\right)+c\left(\sqrt[6]{6x}\right)+d\ln (\sqrt[6]{6x}+1)+e$
$I=\int \frac{dx}{\sqrt{x}+\sqrt[3]{3x}}$
Put $x=u^6,dx=6u^{5}du$
$\int \frac{dx}{\sqrt{x}+\sqrt[3]{3x}}=\int \frac{6u^{5}du}{u^3+u^2}=6\int \frac{u^3}{u+1}du$
$=6\int [u^2-u+1-\frac{1}{u+1}]du$
$=2u^3-3u^2+6u-6\ell n(u+1)+C$
$=2\sqrt{x}-3(\sqrt[3]{3x})+6(\sqrt[6]{6x})-6\ell n(\sqrt[6]{6x}+1)+C$
$\Rightarrow a=2,b=-3,c=6,d=-6$
$\therefore 20a+b+c+d=37$