Question

Integrate $\int {(x^3-1)}^{\frac{1}{3}}{x}^{5}dx$

• A. $\frac{1}{3}\left(\frac{3}{7}\right(x^3-1{)}^{\frac{7}{3}}+\frac{3}{4}(x^3-1{)}^{\frac{4}{3}})$
• B. $\frac{1}{3}\left(\frac{3}{7}\right(x^3-1{)}^{\frac{5}{3}}+\frac{3}{4}(x^3-1{)}^{\frac{4}{3}})$
• C. $\frac{1}{3}\left(\frac{3}{7}\right(x^3-1{)}^{\frac{7}{3}}-\frac{3}{4}(x^3-1{)}^{\frac{4}{3}})$
• D. None of these

Answer 1

The correct option is A $\frac{1}{3}\left(\frac{3}{7}\right(x^3-1{)}^{\frac{7}{3}}+\frac{3}{4}(x^3-1{)}^{\frac{4}{3}})$

${\int }^{}{\left(x^3-1\right)}^{\frac{1}{3}}{x}^{5}dx$

putting $x^3-1=t\Rightarrow x^3=t+1$

$\Rightarrow 3x^{2}dx=dt$

$={\int }^{}{t}^{\frac{1}{3}}\times \frac{\left(t+1\right)}{3}dt$

$=\frac{1}{3}{\int }^{}{t}^{\frac{1}{3}}\left(t+1\right)dt$

$=\frac{1}{3}{\int }^{}\left(t^{\frac{4}{3}}+t^{\frac{1}{3}}\right)dt$

$=\frac{1}{3}\left[\frac{3}{7}{t}^{\frac{7}{3}}+\frac{3}{4}{t}^{\frac{4}{3}}\right]+c$

$=\frac{1}{3}\left[\frac{3}{7}{\left(x^3-1\right)}^{\frac{7}{3}}+\frac{3}{4}{\left(x^3-1\right)}^{\frac{4}{3}}\right]+c$