Question

Evaluate the following

$\int \: \frac{dx}{ \sqrt{x} + \sqrt[3]{x} }$
​

Answer 1

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

Given integral is

$\displaystyle\int\rm \frac{dx}{ \sqrt{x} + \sqrt[3]{x} }$

To evaluate this integral, we have to first remove the fractional exponents.

So, LCM of denominator of fractions in exponents is 6

So, we substitute

$\rm \: x = {y}^{6} \: \: \: \: \bigg[\rm\implies \:y = {\bigg(x\bigg) }^{\dfrac{1}{6} }\bigg ]$

$\rm\implies \:dx = {6y}^{5} \: dy$

So, on substituting these values in above integral, we get

$\rm \: = \: \displaystyle\int\rm \frac{ {6y}^{5} }{ \sqrt{ {y}^{6}} + \sqrt[3]{ {y}^{6} } } \: dy$

$\rm \: = \: 6\displaystyle\int\rm \frac{ {y}^{5} }{ {y}^{3} + {y}^{2} } \: dy$

$\rm \: = \: 6\displaystyle\int\rm \frac{ {y}^{5} }{ {y}^{2}(y + 1) } \: dy$

$\rm \: = \: 6\displaystyle\int\rm \frac{ {y}^{3} }{y + 1} \: dy$

$\rm \: = \: 6\displaystyle\int\rm \frac{ {y}^{3} + 1 - 1}{y + 1} \: dy$

$\rm \: = \: 6\displaystyle\int\rm \bigg[ \frac{ {y}^{3} + 1}{y + 1} - \frac{1}{y + 1} \bigg]\: dy$

$\rm \: = \: 6\displaystyle\int\rm \bigg[ \frac{(y + 1)( {y}^{2} - y + 1) }{y + 1} - \frac{1}{y + 1} \bigg]\: dy$

$\rm \: = \: 6\displaystyle\int\rm \bigg[ {y}^{2} - y + 1 - \frac{1}{y + 1} \bigg]\: dy$

$\rm \: = \: 6\bigg[\dfrac{ {y}^{3} }{3} - \dfrac{ {y}^{2} }{2} + y - log |y + 1| \bigg] + c$

On substituting the value of y, we get

$\rm \: = \: 6\bigg[\dfrac{ \sqrt{x} }{3} - \dfrac{ \sqrt[3]{x} }{2} +{{ \bigg(x\bigg) }^{\dfrac{1}{6}}} - log \bigg|{{ \bigg(x\bigg) }^{\dfrac{1}{6}}} + 1\bigg| \bigg] + c$

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Result Used

$\boxed{\tt{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2}) \: }}$

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

$\boxed{\tt{ \: \displaystyle\int\rm \frac{1}{x} \: dx \: = \: log |x| + c \: }}$

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ADDITIONAL INFORMATION

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

Answer 2

$\sf\red{Solution:-}$

$\sf\longmapsto\boxed{\rm{Answer \: in\: the\: above\: attachment}}$

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@Shivam

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