Finding integral: ∫01ln⁡(1+x)xdx

piarepm

piarepm

Answered

2022-01-03

Finding integral:
01ln(1+x)xdx

Answer & Explanation

Paul Mitchell

Paul Mitchell

Expert

2022-01-04Added 40 answers

Here is an elementary integration of I=01ln(1+x)xdx
I=1201ln(1+x)xdx+3201ln(1+x)xdx=3201ln(1x+x2)xdx
Let J(a)=01ln(12xsina+x2)xdx
J(a)=012cosa(xsina)2+cos2adx=(π2+a)
Then, width J(0)=01ln(1+x2)xdx=12I
I=32J(π6)=32J(0)+0π6J(a)da=34I+320π6(π2+a)da
which leads to
I=π212
sonorous9n

sonorous9n

Expert

2022-01-05Added 34 answers

Hint. One may recall that
ln(1+x)=n=1(1)n1xnn, |x|<1
one may then divide by x and one is allowed to integrate termwise obtaining
01ln(1+x)xdx=n=1(1)n101xn1n=n=1(1)n1n2
Can you take it from here?
user_27qwe

user_27qwe

Expert

2022-01-11Added 230 answers

By integration by parts:
01log(1+x)dxx=01logx1+xdx=201logx1x2dx01
but since 01(logx)xkdx=1(k+1)2, by expanding 11x2 and 11x as geometric series we get:
01log(1+x)dxx=2kodd1k2k11k2=k11k2
=π212

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