Question

If $I=\int (x^{1/3}+(tanx{)}^{1/3})dx=\frac{A}{512}\log \frac{t^4=t^2+1}{(t^2+1{)}^2}+\frac{\sqrt{3}}{2}ta{n}^{-1}\left(\frac{2t^2-1}{2\sqrt{3}}\right)+\frac{3}{4}(ta{n}^{-1}{t}^3{)}^4+C$ where $t^3=tanx$ then A is equal to

• A. 128
• B. 138
• C. 118
• D. 120

Answer 1

The correct option is A 128

Let $I=\int (\sqrt[3]{3x}+\sqrt[3]{3\tan x})dx=I_1+I_2$
Where
$I_1=\int \sqrt[3]{3x}dx=\frac{3x^{4/3}}{4}$
And
$I_2=\int \sqrt[3]{3\tan x}dx$
Put $y=\tan x\Rightarrow dy={\sec }^{2}xdx$
$I_2=\int \frac{\sqrt[3]{3y}}{y^2+1}dy$
Put $u=\sqrt[3]{3y}\Rightarrow du=\frac{1}{3y^{2/3}}dy$
$I_2=3\int \frac{u^3}{u^6+1}du$
Put $s=u^2\Rightarrow ds=2udu$
$I_2=\frac{3}{2}\int \frac{s}{s^3+1}ds=\frac{3}{2}(\int \frac{s+1}{3(s^2-s+1)}ds-\int \frac{1}{3(s+1)}ds)=\frac{1}{2}\int \frac{2s-1}{2(s^2-s+1)}ds+\frac{3}{4}\int \frac{1}{(s^2-s+1)}ds-\frac{1}{2}\int \frac{1}{s+1}=\frac{1}{4}\log (s^2-s+1)-\frac{1}{2}\log (s+1)+\frac{\sqrt{3}}{2}{\tan }^{-1}\left(\frac{2s-1}{\sqrt{3}}\right)$
Hence
$I=\frac{1}{4}\log \left|\frac{t^4-t^2+1}{{(t^2+1)}^2}\right|+\frac{\sqrt{3}}{2}{\tan }^{-1}\left(\frac{2t^2-1}{2\sqrt{3}}\right)+\frac{3}{4}{\left({\tan }^{-1}{t}^3\right)}^4$
Therefore $A=128$