Question

$ \int{(\sqrt[3]{x})}\left( \sqrt[5]{1+\sqrt[3]{{{x}^{4}}}} \right)dx $ is equal to

• A. $ {{\left( 1+{{x}^{\frac{3}{4}}} \right)}^{\frac{5}{6}}}+c $
• B. $ {{\left( 1+{{x}^{\frac{4}{3}}} \right)}^{\frac{5}{6}}}+c $
• C. $ \frac{5}{8}{{\left( 1+{{x}^{\frac{4}{3}}} \right)}^{\frac{6}{5}}}+c $
• D. $ \frac{1}{6}{{\left( 1+{{x}^{\frac{4}{3}}} \right)}^{6}}+c $
• E. $ \frac{15}{8}{{\left( 1+{{x}^{\frac{4}{3}}} \right)}^{\frac{6}{5}}}+c $

Answer 1

Answer: (C)

Let $ I=\int{\sqrt[3]{x}}(\sqrt[5]{1+\sqrt[3]{{{x}^{4}}}})dx $
Put $ \sqrt[3]{{{x}^{4}}}=t $
$ \Rightarrow $ $ \frac{4}{3}.\sqrt[3]{x}dx=dt $
$ \therefore $ $ I=\frac{3}{4}\int{(\sqrt[5]{1+t})dt} $
$=\frac{3}{4}\left[ \frac{{{(1+t)}^{\frac{1}{5}+1}}}{\frac{1}{5}+1} \right]+c $
$=\frac{5}{8}[{{(1+\sqrt[3]{{{x}^{4}}})}^{6/5}}]+c $