Question

$\bigstar \rm \orange{ \quad MIT\ 2006\ Integration\ Bee}$$\dagger \quad \displaystyle \rm \green{\int \dfrac{x^{-\frac{1}{2}}}{1+x^{\frac{1}{3}}}\ dx}$​

Answer 1

Answer:

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$\displaystyle \sf \red{ \int \dfrac{x^{-\frac{1}{2}}}{1+x^{\frac{1}{3}}}\ dx}$

$\longrightarrow \displaystyle \sf \int \dfrac{x^{-\frac{1}{2}}}{1+\left( x^{\frac{1}{6}} \right) ^2}\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{x^{-\frac{5}{6}} .\ x^{\frac{2}{6}} }{1+\left( x^{\frac{1}{6}} \right) ^2}\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{x^{-\frac{5}{6}} .\ \left(x^{\frac{1}{6}} \right)^2 }{1+\left( x^{\frac{1}{6}} \right) ^2}\ dx$

Let's use substitution method ,

➠ $\sf u=x^{\frac{1}{6}}$

➠ $\sf du = \dfrac{1}{6} x^{- \frac{5}{6}}\ dx$

➠ $\sf 6\ du = x^{- \frac{5}{6}}\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{ u^2\ (6\ du)}{1+ u^2}$

$\longrightarrow \displaystyle \sf 6 \int \dfrac{ u^2\ du}{1+ u^2}$

$\longrightarrow \displaystyle \sf 6 \int \dfrac{ u^2 + 1 -1 }{1+ u^2}\ du$

$\longrightarrow \displaystyle \sf 6 \left[ \int du \ - \int \dfrac{ 1 }{1+ u^2}\ du \right]$

$\\ \bullet\ \quad \blue{ \sf \int dx = x + c\ ; \int \dfrac{dx}{1+x^2} = tan^{-1}\ x}$

$\longrightarrow \displaystyle \sf 6 \left[ u \ - tan^{-1}\ u \right]\ + c$

$\bullet\ \qquad \red{\because\ \sf u = x^{\frac{1}{6}}}$

$\longrightarrow \displaystyle \sf \pink{6 \left[\ x^{\frac{1}{6}} \ - tan^{-1}\ \left( x^{\frac{1}{6}} \right)\ \right]\ + c}$

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Answer 2

Explanation:

$\bigstar \rm \orange{ \quad MIT\ 2006\ Integration\ Bee}$$\dagger \quad \displaystyle \rm \green{\int \dfrac{x^{-\frac{1}{2}}}{1+x^{\frac{1}{3}}}\ dx}$

${\displaystyle \int _{a}^{b}(\alpha f+\beta g)(x)\,dx=\alpha \int _{a}^{b}f(x)\,dx+\beta \int _{a}^{b}g(x)\,dx.\,}\int _{a}^{b}(\alpha f+\beta g)(x)\,dx=\alpha \int _{a}^{b}f(x)\,dx+\beta \int _{a}^{b}g(x)\,dx.\,$