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) .](https://www.tinkutara.com/question/Q60051.png)
$${let}\:{use}\:{the}\:{chang}\:\:\mathrm{1}+\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:={t}\:\Rightarrow\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:={t}−\mathrm{1}\:\Rightarrow{x}+\mathrm{1}\:=\left({t}−\mathrm{1}\right)^{\mathrm{3}} \:\Rightarrow \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{0}} \:\:\:\:\frac{{dx}}{\mathrm{1}+^{\mathrm{3}} \sqrt{{x}+\mathrm{1}}}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{\mathrm{3}\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{{t}}\:{dt}\:=\mathrm{3}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}}{{t}}\:{dt} \\ $$$$=\mathrm{3}\:\left\{\:\int_{\mathrm{1}} ^{\mathrm{2}} \:{t}\:{dt}\:−\mathrm{2}\:\int_{\mathrm{1}} ^{\mathrm{2}} {dt}\:+\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{dt}}{{t}}\right\}\:=\mathrm{3}\:\left\{\:\left[\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{2}} \:−\mathrm{2}\:+{ln}\left(\mathrm{2}\right)\right\} \\ $$$$=\mathrm{3}\left\{\:\:\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\:−\mathrm{2}\:\:+{ln}\left(\mathrm{2}\right)\right\}\:=−\frac{\mathrm{3}}{\mathrm{2}}\:+\mathrm{3}{ln}\left(\mathrm{2}\right)\:. \\ $$
Answered by MJS last updated on 16/May/19
![∫_(−1) ^0 (dx/(1+((x+1))^(1/3) ))= [t=1+((x+1))^(1/3) ⇒ dx=3(t−1)^2 dt] =3∫_1 ^2 (((t−1)^2 )/t)dt=3∫_1 ^2 tdt+3∫_1 ^2 (dt/t)−6∫_1 ^2 dt= =[(3/2)t^2 +3ln t −6t]_1 ^2 =−(3/2)+3ln 2](https://www.tinkutara.com/question/Q59967.png)
$$\underset{−\mathrm{1}} {\overset{\mathrm{0}} {\int}}\frac{{dx}}{\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:\Rightarrow\:{dx}=\mathrm{3}\left({t}−\mathrm{1}\right)^{\mathrm{2}} {dt}\right] \\ $$$$=\mathrm{3}\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\frac{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{{t}}{dt}=\mathrm{3}\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{tdt}+\mathrm{3}\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\frac{{dt}}{{t}}−\mathrm{6}\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{dt}= \\ $$$$=\left[\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{2}} +\mathrm{3ln}\:{t}\:−\mathrm{6}{t}\right]_{\mathrm{1}} ^{\mathrm{2}} =−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{3ln}\:\mathrm{2} \\ $$