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Chapter : Integration​

Answer 1

Given Question :-

Evaluate

$\rm \: \displaystyle\int\rm \frac{2}{ {(2 - x)}^{2} } \: \sqrt[3]{\dfrac{2 - x}{2 + x} } \: dx$

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int\rm \frac{2}{ {(2 - x)}^{2} } \: \sqrt[3]{\dfrac{2 - x}{2 + x} } \: dx$

$\rm \: = \: 2\displaystyle\int\rm \frac{1}{ {(2 - x)}^{2} } \: \sqrt[3]{\dfrac{2 - x}{2 + x} } \: dx$

To solve this integral, we use method of Substitution.

So, Substitute

$\rm \: \dfrac{2 + x}{2 - x} = y$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{(2 - x)(1 + 0) - (2 + x)(0 - 1)}{ {(2 - x)}^{2} } = \dfrac{dy}{dx}$

$\rm \: \dfrac{(2 - x) + (2 + x)}{ {(2 - x)}^{2} } = \dfrac{dy}{dx}$

$\rm \: \dfrac{4}{ {(2 - x)}^{2} } = \dfrac{dy}{dx}$

$\rm \: \dfrac{dx}{ {(2 - x)}^{2} } = \dfrac{dy}{4}$

So, on substituting these values, the above integral can be rewritten as

$\rm \: = \: 2 \displaystyle\int\rm \sqrt[3]{ \frac{1}{y} } \: \frac{dy}{4}$

$\rm \: = \: \dfrac{1}{2} \displaystyle\int\rm {\bigg(y \bigg) }^{ \: - \: \dfrac{1}{3} } \: dy \:$

We know,

$\boxed{ \rm{ \: \displaystyle\int\rm {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: }}$

So, using this result, we get

$\rm \: = \: \dfrac{1}{2} \dfrac{{\bigg(y \bigg) }^{ \: - \: \dfrac{1}{3} + 1}}{ - \dfrac{1}{3} + 1} + c$

$\rm \: = \: \dfrac{1}{2} \dfrac{{\bigg(y \bigg) }^{\dfrac{2}{3}}}{ \dfrac{2}{3}} + c$

$\rm \: = \: \dfrac{3}{4}{\bigg(y \bigg) }^{\dfrac{2}{3} } + c$

$\rm \: = \: \dfrac{3}{4}{\bigg(\dfrac{2 + x}{2 - x} \bigg) }^{\dfrac{2}{3} } + c$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$