Question

inigrate the function given above​

Answer 1

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\boxed{\tt \:  ⟼1. \: \tt \:  ⟼ \int {x}^{n} \: dx \: = \dfrac{ {x}^{n + 1} }{n + 1} + c}$

$\boxed{\tt \:  ⟼2. \: \tt \:  ⟼ \int \dfrac{1}{x} \: dx \: = log(x) + c}$

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$\tt \:  ⟼Let \: I = \int \: \dfrac{dx}{ {x}^{ \frac{1}{2} } + {x}^{ \frac{1}{3} } }$

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$\tt \:  ⟼ \: Put \: x = {y}^{6} \: \: (l.c.m \: of \: 2 \: and \: 3)$

$\tt \:  ⟼ \: \dfrac{d}{dx}x \: = \dfrac{d}{dx} {y}^{6}$

$\tt \:  ⟼ \: 1 = {6y}^{5} \dfrac{dy}{dx}$

$\tt \:  ⟼dx \: = {6y}^{5} dy$

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$\tt\implies \: \: I = \tt \:  ⟼ \int\dfrac{ {6y}^{5}dy }{ {y}^{3} + {y}^{2} }$

$\tt\implies \: \: I = \tt \:  ⟼ 6\int\dfrac{ {y}^{5}dy }{ {y}^{2} (y + 1)}$

$\tt\implies \: \: I = \tt \:  ⟼6 \int\dfrac{ {y}^{3} \: dy}{y + 1}$

$\tt\implies \: \: I = \tt \:  ⟼6 \int\dfrac{ {y}^{3} + 1 - 1 }{y + 1} \: \: dy$

$\tt\implies \: \: I = \tt \:  ⟼6 \int\dfrac{(y + 1)( {y}^{2} + 1 - y) - 1}{y + 1} \: \: dy$

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

$\tt\implies \: \: I = \tt \:  ⟼ 6\int\ \bigg( {y}^{2} + 1 - y - \dfrac{1}{y + 1} \bigg)dy$

$\tt \:  ⟼ \: I = \dfrac{ {y}^{3} }{3} + y - \dfrac{ {y}^{2} }{2} - log(1 + y) + c$

$\tt \:  ⟼on \: substituting \: y \: = {x}^{ \frac{1}{6} } we \: get$

$\tt \:  ⟼ \: I = \dfrac{ {x}^{ \frac{1}{2} } }{3} + {x}^{ \frac{1}{6} } - \dfrac{ {x}^{ \frac{1}{3} } }{2} - log(1 + {x}^{ \frac{1}{6} } ) + c$

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