Question

$\displaystyle \bf \red{ \int_{0}^{1} \bigg \{ \frac{1}{ \sqrt[k]{x} } \bigg \} \: dx}$​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle \bf{ \int_{0}^{1} \bigg \{ \frac{1}{ \sqrt[k]{x} } \bigg \} \: dx}$

can be rewritten as

$\rm \:  =  \: \:\displaystyle \bf{ \int_{0}^{1} \frac{1}{ {\bigg(x\bigg) }^{ \dfrac{1}{k} } } \: dx}$

$\rm \:  =  \: \displaystyle \bf{ \int_{0}^{1}} \: {\bigg(x\bigg) }^{ - \dfrac{1}{k} } \: dx$

$\rm \:  =  \: \dfrac{{\bigg(x\bigg) }^{ - \dfrac{1}{k} + 1 }}{ - \dfrac{1}{k} + 1 } \bigg| _{0}^{1}$

$\rm \:  =  \: \dfrac{{\bigg(x\bigg) }^{\dfrac{k - 1}{k}}}{ \dfrac{k - 1}{k} } \bigg| _{0}^{1}$

$\rm \:  =  \: \dfrac{k}{k - 1} (1 - 0)$

$\rm \:  =  \: \dfrac{k}{k - 1}$

Hence,

$\rm\implies \:\:\boxed{\tt{ \displaystyle \bf{ \int_{0}^{1} \bigg \{ \frac{1}{ \sqrt[k]{x} } \bigg \} \: dx} = \frac{k}{k - 1} \: }}$

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MORE TO KNOW

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

Answer 2

Answer:

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

Given integral is

$\rm :\longmapsto\:\displaystyle \bf{ \int_{0}^{1} \bigg \{ \frac{1}{ \sqrt[k]{x} } \bigg \} \: dx}$

can be rewritten as

$\rm \:  =  \: \:\displaystyle \bf{ \int_{0}^{1} \frac{1}{ {\bigg(x\bigg) }^{ \dfrac{1}{k} } } \: dx}$

$\rm \:  =  \: \displaystyle \bf{ \int_{0}^{1}} \: {\bigg(x\bigg) }^{ - \dfrac{1}{k} } \: dx$

$\rm \:  =  \: \dfrac{{\bigg(x\bigg) }^{ - \dfrac{1}{k} + 1 }}{ - \dfrac{1}{k} + 1 } \bigg| _{0}^{1}$

$\rm \:  =  \: \dfrac{{\bigg(x\bigg) }^{\dfrac{k - 1}{k}}}{ \dfrac{k - 1}{k} } \bigg| _{0}^{1}$

$\rm \:  =  \: \dfrac{k}{k - 1} (1 - 0)$

$\rm \:  =  \: \dfrac{k}{k - 1}$

Hence,

$\rm\implies \:\:\boxed{\tt{ \displaystyle \bf{ \int_{0}^{1} \bigg \{ \frac{1}{ \sqrt[k]{x} } \bigg \} \: dx} = \frac{k}{k - 1} \: }}$

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