Nebo - X - Page 1
10ln(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

π40ln(tant+1)tan2t+1·sec2tdt=π40ln(tant+1)

So now, how can we solve this integral?

If we have an integral in general we have following identity

a0f(x)dx=a0f(a-x)dx

which can be shown by changing variable x= a-u and rename the variable to × again

π40ln(tant+1)=π40ln(tan(π4-t)+1)

Now, we have

tan(π4-t)=tanπ4-tant1+tanπ4·tant=1-tant1+tant

we can
do the
summation

π40ln(1-tant1+tant+1)dt=π40ln(21+tant)dt=π40ln(2)dt-π/40ln(tant+1)dt
π/40ln(tant+1)dt=10ln(x+1)x2+1dx

Remember this was
the integral we started
with so finally

10ln(x+1)x2+1dx=ln(2)·π42=ln2.π8

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