Question

$\int \left(e^{\sqrt[3]{3x}}dx\right)=$

• A. $x^{2/3}-2x^{1/3}+2+c$
• B. $(x^{2/3}-2x^{1/3}+2)\exp \left(\sqrt[3]{3x}\right)+c$
• C. $3(x^{2/3}-2x^{1/3}+2)\exp \left(\sqrt[3]{3x}\right)+c$
• D. $2(x^{2/3}-2x^{1/3}+2)\exp \left(\sqrt[3]{3x}\right)+c$

Answer 1

The correct option is C $3(x^{2/3}-2x^{1/3}+2)\exp \left(\sqrt[3]{3x}\right)+c$

$I=\int e^{\sqrt[3]{3x}}dx$

Put $\sqrt[3]{3x}=t$

$\frac{1}{3}{x}^{-2/3}dx=dt$

$dx=\frac{3}{t^{-2}}dt$

So, $I=3\int {t^{2}e}^{t}dt$

$=3[{t^{2}e}^t-2\int t{e}^{t}dt]+c$

$=3{t^{2}e}^t-6[t{e}^t-\int e^{t}dt]+c$

$=3{t^{2}e}^t-6t{e}^t+6e^t+c$

$=3(x^{2/3}-2x^{1/3}+2)e^{\sqrt[3]{3x}}+c$