1∫0ln(x+1)x2+1dx=?
we substitute
x=tant=sintcost
dx=cost·cost+sint·sintcos2t=1cos2t=sec2tdt
Also
note that
tan2t+1=sin2tcos2t+1=sec2t
After substituting new variable and change integral bounds accordingly we have
π4∫0ln(tant+1)tan2t+1·sec2tdt=π4∫0ln(tant+1)
So now, how can we solve this integral?
If we have an integral in general we have following identity
a∫0f(x)dx=a∫0f(a-x)dx
which can be shown by changing variable x= a-u and rename the variable to × again
π4∫0ln(tant+1)=π4∫0ln(tan(π4-t)+1)
Now, we have
tan(π4-t)=tanπ4-tant1+tanπ4·tant=1-tant1+tant
we can
do the
summation
π4∫0ln(1-tant1+tant+1)dt=π4∫0ln(21+tant)dt=π4∫0ln(2)dt-π/4∫0ln(tant+1)dt
π/4∫0ln(tant+1)dt=1∫0ln(x+1)x2+1dx
Remember this was
the integral we started
with so finally
1∫0ln(x+1)x2+1dx=ln(2)·π42=ln2.π8
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