Integral of (ln(3x-1))/(3x-1) dx
The solution
Detail solution
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Let u=3x−1.
Then let du=3dx and substitute 3du:
∫3ulog(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫ulog(u)du=3∫ulog(u)du
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There are multiple ways to do this integral.
Method #1
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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
Method #2
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Let u=log(u).
Then let du=udu 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(u)2
So, the result is: 6log(u)2
Now substitute u back in:
6log(3x−1)2
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Now simplify:
6log(3x−1)2
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Add the constant of integration:
6log(3x−1)2+constant
The answer is:
6log(3x−1)2+constant
The answer (Indefinite)
[src]
/
| 2
| log(3*x - 1) log (3*x - 1)
| ------------ dx = C + -------------
| 3*x - 1 6
|
/
∫3x−1log(3x−1)dx=C+6log(3x−1)2
(-283.773849354205 - 1.68049929284057j)
(-283.773849354205 - 1.68049929284057j)
Use the examples entering the upper and lower limits of integration.