If ∫ d x x √ 1 − x 3 = a l n ( √ 1 − x 3 + b − √ 1 − x 3 + 1 ) + k , then Options: A . &nbsp b = 1 , a = 1 B . &nbsp b = 1 , a = − 1 3 C . &nbsp b = 1 , a = − 2 3 D . &nbsp None of these

Answer: Option B : B I = ∫ d x x √ 1 − x 3 P u t 1 − x 3 = t 2 3 x 2 d x = − 2 t d t I = − 2 3 ∫ d t ( 1 − t 2 ) = − 1 3 ∫ ( 1 1 − t + 1 1 + t ) d t . = − 1 3 l o g 1 + t 1 − t + c = − 1 3 l o g 1 + √ 1 − x 3 1 − √ 1 − x 3 + c .

If ∫dxx√1−x3=aln(√1−x3+b−√1&#

Discussion Forum : Indefinite Integration

Question -

If $\int \frac{d x}{x \sqrt{1 - x^{3}}} = a l n (\frac{\sqrt{1 - x^{3}} + b}{- \sqrt{1 - x^{3}} + 1}) + k$ , then

Options:

A .

b = 1, a = 1

B .

b = 1, a = - \frac{1}{3}

C .

b = 1, a = - \frac{2}{3}

D . None of these

Answer: Option B
:
B

I = \int \frac{d x}{x \sqrt{1 - x^{3}}} P u t 1 - x^{3} = t^{2} 3 x^{2} d x = - 2 t d t I = \frac{- 2}{3} \int \frac{d t}{(1 - t^{2})} = \frac{- 1}{3} \int (\frac{1}{1 - t} + \frac{1}{1 + t}) d t . = \frac{- 1}{3} l o g \frac{1 + t}{1 - t} + c = \frac{- 1}{3} l o g \frac{1 + \sqrt{1 - x^{3}}}{1 - \sqrt{1 - x^{3}}} + c .

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