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Question Number 32301 by abdo imad last updated on 22/Mar/18

calculate ∫_1 ^e ln(1+(√x))dx .

calculate∫1eln(1+x)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) .

ch.x=tgiveI=∫1e(2t)ln(1+t)dt.bypartsI=([t2ln(1+t)]1e−∫1et21+tdt)=(eln(1+e)−ln(2))−∫1et2−1+11+tdt∫1et2−1+11+tdt=∫1e(t−1)dt+∫1edt1+t=[t22−t]1e+[ln(1+t)]1e=e2−e−12+1+ln(1+e)−ln(2)=e2−e+12+ln(1+e)−ln(2)I=eln(1+e)−ln(2)−e2+e−12−ln(1+e)+ln(2)I=(e−1)ln(1+e)+e−e2−12.

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