If ∫ d x x √ 1 − x 3 = a l o g ∣ ∣ ∣ √ 1 − x 3 − 1 √ 1 − x 3 + 1 ∣ ∣ ∣ + C then a = Options: A . &nbsp 1 3 B . &nbsp 2 3 C . &nbsp − 1 3 D . &nbsp − 2 3

Answer: Option A : A Put<br> 1 − x 3 = t 2 ⇒ − 3 x 2 d x = 2 t d t ∴ ∫ d x x √ 1 − x 3 = ∫ x 2 x 3 √ 1 − x 3 d x = 2 3 ∫ d t t 2 − 1 = 1 3 l o g ∣ ∣ t − 1 t + 1 ∣ ∣ + C = 1 3 l o g ∣ ∣ ∣ √ 1 − x 3 − 1 √ 1 − x 3 + 1 ∣ ∣ ∣ + C ∴ a = 1 3

If ∫dxx√1−x3=a log∣∣∣√1&

Discussion Forum : Indefinite Integration

Question -

If $\int \frac{d x}{x \sqrt{1 - x^{3}}} = a l o g ∣ ∣ ∣ \frac{\sqrt{1 - x^{3}} - 1}{\sqrt{1 - x^{3}} + 1} ∣ ∣ ∣ + C$ then a =

Options:

A .

\frac{1}{3}

B .

\frac{2}{3}

C .

- \frac{1}{3}

D .

- \frac{2}{3}

Answer: Option A
:
A

Put<br> $1 - x^{3} = t^{2} \Rightarrow - 3 x^{2} d x = 2 t d t ∴ \int \frac{d x}{x \sqrt{1 - x^{3}}} = \int \frac{x^{2}}{x^{3} \sqrt{1 - x^{3}}} d x = \frac{2}{3} \int \frac{d t}{t^{2} - 1} = \frac{1}{3} l o g ∣ ∣ \frac{t - 1}{t + 1} ∣ ∣ + C = \frac{1}{3} l o g ∣ ∣ ∣ \frac{\sqrt{1 - x^{3}} - 1}{\sqrt{1 - x^{3}} + 1} ∣ ∣ ∣ + C ∴ a = \frac{1}{3}$

Was this answer helpful ?

Next Question

Discussion Forum : Indefinite Integration

Submit Your Solution hear: