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Q)

Integrate the function $(x^3-1)^\frac{1}{3}\;x^5$

$\begin{array}{1 1}\frac{1}{7}(x^3-1)^\frac{7}{3}+\frac{1}{4}(x^3-1)^\frac{4}{3}+c \\ \frac{1}{5}(x^3-1)^\frac{5}{3}+\frac{1}{4}(x^3-1)^\frac{4}{3}+c \\ \frac{1}{7}(x^2-1)^\frac{7}{3}+\frac{1}{4}(x^2-1)^\frac{4}{3}+c\\ \frac{1}{5}(x^2-1)^\frac{5}{3}+\frac{1}{4}(x^2-1)^\frac{4}{3}+c\end{array}$

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A)
Toolbox:
• Method of substitution :
• Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
• Consider $I=\int f(x)dx.$
• Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
• dx=g'(t)dt.
• Thus $I=\int f(g(t).g'(t))dt.$
Given $I=\int(x^3-1).x^5$.

This can be written as

$\;\;\;=\int (x^3-1)\frac{1}{3}.x^3.x^2dx$.

Let $x^3-1=t\Rightarrow x^3=(t+1).$

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

Now substituting for x and dx we get,

$I=\int (t)^\frac{1}{3}.(t+1).\frac{dt}{3}$.

\;\;\;=\frac{1}{3}\int(t^\frac{4}{3}+t^\frac{1}{3})dt.$On integrating we get,$\;\;\;=\frac{1}{3}\begin{bmatrix}\frac{t^\frac{7}{3}}{\frac{7}{3}}+\frac{t^\frac{4}{3}}{\frac{4}{3}}\end{bmatrix}+c.\;\;\;=\frac{1}{3}[\frac{3}{7}(t^\frac{7}{3})+\frac{3}{4}(t^\frac{4}{3})]+c.$Substituting back for t we get,$\;\;\;=\frac{1}{7}(x^3-1)^\frac{7}{3}+\frac{1}{4}(x^3-1)^\frac{4}{3}+c.$Hence$\int(x^3-1).x^5=\frac{1}{7}(x^3-1)^\frac{7}{3}+\frac{1}{4}(x^3-1)^\frac{4}{3}+c.\$