Integral of (1+ln(x-1))/(x-1) dx
The solution
Detail solution
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There are multiple ways to do this integral.
Method #1
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Let u=log(x−1)+1.
Then let du=x−1dx and substitute du:
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The integral of un is n+1un+1 when n=−1:
∫udu=2u2
Now substitute u back in:
2(log(x−1)+1)2
Method #2
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Rewrite the integrand:
x−1log(x−1)+1=x−1log(x−1)+x−11
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Integrate term-by-term:
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Let u=x−1.
Then let du=dx and substitute du:
∫ulog(u)du
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Let u=u1.
Then let du=−u2du and substitute −du:
∫(−ulog(u1))du
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The integral of a constant times a function is the constant times the integral of the function:
∫ulog(u1)du=−∫ulog(u1)du
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Let u=log(u1).
Then let du=−udu and substitute −du:
∫(−u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫udu=−∫udu
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The integral of un is n+1un+1 when n=−1:
∫udu=2u2
So, the result is: −2u2
Now substitute u back in:
−2log(u1)2
So, the result is: 2log(u1)2
Now substitute u back in:
2log(u)2
Now substitute u back in:
2log(x−1)2
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Let u=x−1.
Then let du=dx and substitute du:
∫u1du
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The integral of u1 is log(u).
Now substitute u back in:
log(x−1)
The result is: 2log(x−1)2+log(x−1)
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Now simplify:
2(log(x−1)+1)2
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Add the constant of integration:
2(log(x−1)+1)2+constant
The answer is:
2(log(x−1)+1)2+constant
The answer (Indefinite)
[src]
/
| 2
| 1 + log(x - 1) (1 + log(x - 1))
| -------------- dx = C + -----------------
| x - 1 2
|
/
∫x−1log(x−1)+1dx=C+2(log(x−1)+1)2
2
(pi*I + log(2))
oo - ---------------- - pi*I
2
∞−iπ−2(log(2)+iπ)2
=
2
(pi*I + log(2))
oo - ---------------- - pi*I
2
∞−iπ−2(log(2)+iπ)2
oo - (pi*i + log(2))^2/2 - pi*i
(897.312398743146 - 138.514234126527j)
(897.312398743146 - 138.514234126527j)
Use the examples entering the upper and lower limits of integration.