Question Number 32302 by abdo imad last updated on 22/Mar/18
$${calculate}\:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\frac{{dx}}{{x}\:+{x}\sqrt{{x}}}\:. \\ $$
Answered by sma3l2996 last updated on 23/Mar/18
![let t=(√x)⇒dx=2tdt A=∫_1 ^2 (dx/(x+x(√x)))=2∫_1 ^(√2) ((tdt)/(t^2 (1+t)))=2∫_1 ^(√2) (dt/(t(1+t)))=2∫_1 ^(√2) ((1/t)−(1/(t+1)))dt =2[ln∣(t/(t+1))∣]_1 ^(√2) =2(ln(((√2)/( (√2)+1)))+ln2)](https://www.tinkutara.com/question/Q32327.png)
$${let}\:{t}=\sqrt{{x}}\Rightarrow{dx}=\mathrm{2}{tdt} \\ $$$${A}=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{dx}}{{x}+{x}\sqrt{{x}}}=\mathrm{2}\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \frac{{tdt}}{{t}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)}=\mathrm{2}\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \frac{{dt}}{{t}\left(\mathrm{1}+{t}\right)}=\mathrm{2}\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \left(\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt} \\ $$$$=\mathrm{2}\left[{ln}\mid\frac{{t}}{{t}+\mathrm{1}}\mid\right]_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} =\mathrm{2}\left({ln}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}\right)+{ln}\mathrm{2}\right) \\ $$