Question
Integrate the functions in Exercises:
$(\text{x}^3-1)^{\frac{1}{3}}\text{ x}^5$
$(\text{x}^3-1)^{\frac{1}{3}}\text{ x}^5$
✓
Answer
$\text{Let I}=\int(\text{x}^3 - 1)^{\frac{1}{3}}\text{x}^5\text{ dx}$ $=\int(\text{x}^3 - 1\big)^{\frac{1}{3}}\text{x}^3\text{x}^2\text{ dx}=\frac{1}{3}\int(\text{x}^3 - 1)^{\frac{1}{3}}\text{ x}^3(3\text{x}^2\text{ dx})\ \ \ \ \ \ .....\text{(i)}$ Putting $\text{ x}^3-1=\text{t}\ \ \ \ \Rightarrow \ \ \ \ \ \text{ x}^3=\text{t}+1\ \ \ \Rightarrow \ \ \ \ \text{3x}^2=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow \ \ \ \ 3\text{x}^2\text{ dx}=\text{dt} $ From eq. (i), $\text{I}=\frac{1}{3}\int\text{t}^{\frac{1}{3}}(\text{t}+1)\text{ dt}$ $=\frac{1}{3}\int\bigg(\text{t}^{\frac{4}{3}}+\text{t}^{\frac{1}{3}}\bigg)\text{ dt}=\frac{1}{3}\bigg(\int\text{t}^{\frac{4} {3}}\text{ dt}+\int\text{t}^{\frac{1}{3}}\text{ dt}\bigg) $$=\frac{1}{3}\Bigg(\frac{\text{t}^{\frac{7}{3}}}{\frac{7}{3}}+\frac{\text{t}^{\frac{4}{3}}}{\frac{4}{3}}\Bigg)+\text{ c} =\frac{1}{3}\bigg(\frac{3}{7}\text{ t}^{\frac{7}{3}}+\frac{3}{4}\text{ t}^{\frac{4}{3}}\bigg)+\text{c}$
$=\frac{1}{7}\text{t}^{\frac{7}{3}}+\frac{1}{4}\text{t}^{\frac{4}{3}}+ \text{c} $ $=\frac{1}{7}(\text{x}^3 - 1)^{\frac{7}{3}}+\frac{1}{4}(\text{x}^3 - 1)^{\frac{4}{3}}+\text{c} $
$=\frac{1}{7}\text{t}^{\frac{7}{3}}+\frac{1}{4}\text{t}^{\frac{4}{3}}+ \text{c} $ $=\frac{1}{7}(\text{x}^3 - 1)^{\frac{7}{3}}+\frac{1}{4}(\text{x}^3 - 1)^{\frac{4}{3}}+\text{c} $
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