dx-x-2-3-1-x-2-3-

Question Number 36597 by rahul 19 last updated on 03/Jun/18

\int \frac{d x}{x^{\frac{2}{3}} (1 + x^{\frac{2}{3}})} = ?

Commented by rahul 19 last updated on 03/Jun/18

I tried by taking x^{\frac{2}{3}} c o m m o n f r o m

t h e b r a c k e t o f d e n o m i n a t o r a n d

t h e n (x^{\frac{- 2}{3}} + 1) = t b u t {c o u l d n}^{'} t reach till

end . Can someone try like this ?

Commented by abdo mathsup 649 cc last updated on 03/Jun/18

c h a n g e m e n t x^{\frac{2}{3}} = t g i v e x = t^{\frac{3}{2}}

I = \int \frac{1}{t (1 + t)} \frac{3}{2} t^{\frac{1}{2}} d t

= \frac{3}{2} \int \frac{\sqrt{t}}{t (1 + t)} d t a n d c h a n g e m e n t \sqrt{t} = x g i v e

I = \frac{3}{2} \int \frac{x}{x^{2} (1 + x^{2})} 2 x d x

= 3 \int \frac{d x}{1 + x^{2}} = 3 a r c t a n x + c

= 3 a r c t a n (t \sqrt{t}) + c .

Answered by Joel579 last updated on 03/Jun/18

I = \int \frac{d x}{x^{2 / 3} (1 + x^{2 / 3})} (t = x^{1 / 3} \to d t = \frac{x^{- 2 / 3}}{3} d x \Leftrightarrow d x = 3 t^{2} d t)

= \int \frac{3 t^{2}}{t^{2} (1 + t^{2})} d t

= 3 \tan^{- 1} (t) + C

= 3 \tan^{- 1} (x^{1 / 3}) + C

Answered by sma3l2996 last updated on 03/Jun/18

l e t t = x^{1 / 3} \Rightarrow d t = \frac{d x}{3 x^{2 / 3}}

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

\int \frac{d x}{x^{2 / 3} (1 + x^{2 / 3})} = 3 \int \frac{d t}{1 + t^{2}} = 3 {t a n}^{- 1} (t) + C

= 3 {t a n}^{- 1} (x^{\frac{1}{3}}) + C