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Integral of log(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).
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:
2log(x+1)2
Method #2
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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:
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The integral of a constant times a function is the constant times the integral of the function:
∫(−u)du=−∫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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Now simplify:
2log(x+1)2
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Add the constant of integration:
2log(x+1)2+constant
The answer is:
2log(x+1)2+constant
The answer (Indefinite)
[src]
/
| 2
| log(x + 1) log (x + 1)
| ---------- dx = C + -----------
| x + 1 2
|
/
2(log(x+1))2
The graph
2(log2)2
=
2log(2)2
Use the examples entering the upper and lower limits of integration.