∫ √ 1 + 3 √ x 3 √ x 2 d x is equal to Options: A . &nbsp ( 1 + x 1 / 3 ) 3 / 2 + C B . &nbsp − ( 1 + x 1 / 3 ) 3 / 2 + C C . &nbsp 2 ( 1 + x 1 / 3 ) 3 / 2 + C D . &nbsp None of these

Answer: Option C : C ∫ √ 1 + 3 √ x 3 √ x 2 d x = ∫ x − 2 3 ( 1 + x 1 3 ) 1 / 2 d x = ∫ x − 2 / 3 ( 1 + x 1 / 3 ) 1 / 2 d x 1 + x 1 / 3 = t 2 x − 2 / 3 d x = 6 t d t = ∫ 6 t 2 d t = 2 t 3 + c = 2 ( 1 + x 3 ) 3 / 2 + c

∫√1+3√x3√x2dx is equal to

Discussion Forum : Indefinite Integration

Question -

$\int \frac{\sqrt{1 + \sqrt[3]{x}}}{\sqrt[3]{x^{2}}} d x$ is equal to

Options:

A .

(1 + x^{1 / 3})^{3 / 2} + C

B .

- (1 + x^{1 / 3})^{3 / 2} + C

C .

2 (1 + x^{1 / 3})^{3 / 2} + C

D . None of these

Answer: Option C
:
C

\int \frac{\sqrt{1 + \sqrt[3]{x}}}{\sqrt[3]{x^{2}}} d x = \int x^{- \frac{2}{3}} {(1 + x^{\frac{1}{3}})}^{1 / 2} d x

= \int x^{- 2 / 3} {(1 + x^{1 / 3})}^{1 / 2} d x

1 + x^{1 / 3} = t^{2}

x^{- 2 / 3} d x = 6 t d t

= \int 6 t^{2} d t

= 2 t^{3} + c

= 2 {(1 + x^{3})}^{3 / 2} + c

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Discussion Forum : Indefinite Integration

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