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Question Number 59946 by Sardor2211 last updated on 16/May/19

Commented by maxmathsup by imad last updated on 17/May/19
$let use the chang 1+(x+1)^(1/3) =t ⇒(x+1)^(1/3) =t−1 ⇒x+1 =(t−1)^3 ⇒ ∫_(−1) ^0 (dx/(1+^3 (√(x+1)))) =∫_1 ^2 ((3(t−1)^2 )/t) dt =3 ∫_1 ^2 ((t^2 −2t+1)/t) dt =3 { ∫_1 ^2 t dt −2 ∫_1 ^2 dt +∫_1 ^2 (dt/t)} =3 { [(t^2 /2)]_1 ^2 −2 +ln(2)} =3{ 2−(1/2) −2 +ln(2)} =−(3/2) +3ln(2) .$
$l e t u s e t h e c h a n g 1 + {(x + 1)}^{\frac{1}{3}} = t \Rightarrow {(x + 1)}^{\frac{1}{3}} = t - 1 \Rightarrow x + 1 = {(t - 1)}^{3} \Rightarrow$ $\int_{- 1}^{0} \frac{d x}{1 +^{3} \sqrt{x + 1}} = \int_{1}^{2} \frac{3 {(t - 1)}^{2}}{t} d t = 3 \int_{1}^{2} \frac{t^{2} - 2 t + 1}{t} d t$ $= 3 {\int_{1}^{2} t d t - 2 \int_{1}^{2} d t + \int_{1}^{2} \frac{d t}{t}} = 3 {{[\frac{t^{2}}{2}]}_{1}^{2} - 2 + l n (2)}$ $= 3 {2 - \frac{1}{2} - 2 + l n (2)} = - \frac{3}{2} + 3 l n (2) .$
	Answered by MJS last updated on 16/May/19

	$\underset{- 1}{\int^{0}} \frac{d x}{1 + \sqrt[3]{x + 1}} =$ $[t = 1 + \sqrt[3]{x + 1} \Rightarrow d x = 3 {(t - 1)}^{2} d t]$ $= 3 \underset{1}{\int^{2}} \frac{{(t - 1)}^{2}}{t} d t = 3 \underset{1}{\int^{2}} t d t + 3 \underset{1}{\int^{2}} \frac{d t}{t} - 6 \underset{1}{\int^{2}} d t =$ $= {[\frac{3}{2} t^{2} + 3 \ln t - 6 t]}_{1}^{2} = - \frac{3}{2} + 3 \ln 2$
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