(C) Let $I = \int \frac{d x}{x^{2 / 3}(1+x^{2 / 3})}$.Substitute $x^{1/3} = t$. Then $x = t^3$,which implies $dx = 3t^2 dt$.Substituting these into th

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$3 \operatorname{Tan}^{-1}\left(x^{1 / 3}\right)+c$

(C) Let $I = \int \frac{d x}{x^{2 / 3}(1+x^{2 / 3})}$.Substitute $x^{1/3} = t$. Then $x = t^3$,which implies $dx = 3t^2 dt$.Substituting these into the integral:$I = \int \frac{3t^2 dt}{(t^2)(1+t^2)} = \int \frac{3 dt}{1+t^2}$.The integral of $\frac{1}{1+t^2}$ is $\tan^{-1}(t)$.Therefore,$I = 3 \tan^{-1}(t) + c$.Substituting back $t = x^{1/3}$,we get $I = 3 \tan^{-1}(x^{1/3}) + c$.

Class 12 Mathematics 7-1.Indefinite Integral Integration by substitution

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$\int \frac{d x}{x^{2 / 3}\left(1+x^{2 / 3}\right)}=$

AP EAMCET 2017 All AP EAMCET

A
$3 \operatorname{Sin}^{-1}\left(x^{1 / 3}\right)+c$
B
$3 \operatorname{Cos}^{-1}\left(x^{1 / 3}\right)+c$
C
$3 \operatorname{Tan}^{-1}\left(x^{1 / 3}\right)+c$
D
$3 \operatorname{Sec}^{-1}\left(x^{1 / 3}\right)+c$

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7-1.Indefinite Integral

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