Question Number 32301 by abdo imad last updated on 22/Mar/18
$${calculate}\:\int_{\mathrm{1}} ^{{e}} \:{ln}\left(\mathrm{1}+\sqrt{{x}}\right){dx}\:. \\ $$
Commented by abdo imad last updated on 01/Apr/18
![ch. (√x) =t give I = ∫_1 ^(√e) (2t) ln(1+t)dt .by parts I =( [ t^2 ln(1+t)]_1 ^(√e) − ∫_1 ^(√e) (t^2 /(1+t))dt) =( e ln(1+(√e)) −ln(2) ) − ∫_1 ^(√e) ((t^2 −1+1)/(1+t))dt ∫_1 ^(√e) ((t^2 −1+1)/(1+t)) dt = ∫_1 ^(√e) (t−1) dt +∫_1 ^(√e) (dt/(1+t)) =[(t^2 /2) −t]_1 ^(√e) +[ln(1+t)]_1 ^(√e) =(e/2) −(√e) −(1/2) +1 +ln(1+(√e)) −ln(2) =(e/2) −(√e) +(1/2) +ln(1+(√e)) −ln(2) I =e ln(1+(√e)) −ln(2) −(e/2) +(√e) −(1/2) −ln(1+(√e)) +ln(2) I =(e−1)ln(1+(√e)) +(√e) −(e/2) −(1/2) .](https://www.tinkutara.com/question/Q32753.png)
$${ch}.\:\sqrt{{x}}\:={t}\:{give}\:{I}\:=\:\int_{\mathrm{1}} ^{\sqrt{{e}}} \left(\mathrm{2}{t}\right)\:{ln}\left(\mathrm{1}+{t}\right){dt}\:.{by}\:{parts} \\ $$$${I}\:=\left(\:\left[\:{t}^{\mathrm{2}} {ln}\left(\mathrm{1}+{t}\right)\right]_{\mathrm{1}} ^{\sqrt{{e}}} \:\:−\:\int_{\mathrm{1}} ^{\sqrt{{e}}} \:\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}}{dt}\right) \\ $$$$=\left(\:{e}\:{ln}\left(\mathrm{1}+\sqrt{{e}}\right)\:−{ln}\left(\mathrm{2}\right)\:\right)\:−\:\int_{\mathrm{1}} ^{\sqrt{{e}}} \:\:\frac{{t}^{\mathrm{2}} \:−\mathrm{1}+\mathrm{1}}{\mathrm{1}+{t}}{dt} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{{e}}} \:\frac{{t}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{\mathrm{1}+{t}}\:{dt}\:=\:\int_{\mathrm{1}} ^{\sqrt{{e}}} \left({t}−\mathrm{1}\right)\:{dt}\:\:+\int_{\mathrm{1}} ^{\sqrt{{e}}} \:\:\frac{{dt}}{\mathrm{1}+{t}} \\ $$$$=\left[\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:−{t}\right]_{\mathrm{1}} ^{\sqrt{{e}}} \:\:+\left[{ln}\left(\mathrm{1}+{t}\right)\right]_{\mathrm{1}} ^{\sqrt{{e}}} \\ $$$$=\frac{{e}}{\mathrm{2}}\:−\sqrt{{e}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{1}\:\:+{ln}\left(\mathrm{1}+\sqrt{{e}}\right)\:−{ln}\left(\mathrm{2}\right) \\ $$$$=\frac{{e}}{\mathrm{2}}\:−\sqrt{{e}}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:+{ln}\left(\mathrm{1}+\sqrt{{e}}\right)\:−{ln}\left(\mathrm{2}\right) \\ $$$${I}\:={e}\:{ln}\left(\mathrm{1}+\sqrt{{e}}\right)\:−{ln}\left(\mathrm{2}\right)\:−\frac{{e}}{\mathrm{2}}\:+\sqrt{{e}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:−{ln}\left(\mathrm{1}+\sqrt{{e}}\right)\:+{ln}\left(\mathrm{2}\right) \\ $$$${I}\:=\left({e}−\mathrm{1}\right){ln}\left(\mathrm{1}+\sqrt{{e}}\right)\:+\sqrt{{e}}\:−\frac{{e}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$