Integral of ln(3x+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=3x+1.
Then let du=3dx and substitute 3du:
∫9log(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫3log(u)du=3∫log(u)du
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Use integration by parts:
∫udv=uv−∫vdu
Let u(u)=log(u) and let dv(u)=1.
Then du(u)=u1.
To find v(u):
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The integral of a constant is the constant times the variable of integration:
∫1du=u
Now evaluate the sub-integral.
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The integral of a constant is the constant times the variable of integration:
∫1du=u
So, the result is: 3ulog(u)−3u
Now substitute u back in:
−x+3(3x+1)log(3x+1)−31
Method #2
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Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=log(3x+1) and let dv(x)=1.
Then du(x)=3x+13.
To find v(x):
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The integral of a constant is the constant times the variable of integration:
∫1dx=x
Now evaluate the sub-integral.
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The integral of a constant times a function is the constant times the integral of the function:
∫3x+13xdx=3∫3x+1xdx
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Rewrite the integrand:
3x+1x=31−3⋅(3x+1)1
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Integrate term-by-term:
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The integral of a constant is the constant times the variable of integration:
∫31dx=3x
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The integral of a constant times a function is the constant times the integral of the function:
∫(−3⋅(3x+1)1)dx=−3∫3x+11dx
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Let u=3x+1.
Then let du=3dx and substitute 3du:
∫9u1du
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The integral of a constant times a function is the constant times the integral of the function:
∫3u1du=3∫u1du
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The integral of u1 is log(u).
So, the result is: 3log(u)
Now substitute u back in:
3log(3x+1)
So, the result is: −9log(3x+1)
The result is: 3x−9log(3x+1)
So, the result is: x−3log(3x+1)
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Now simplify:
−x+3(3x+1)log(3x+1)−31
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Add the constant of integration:
−x+3(3x+1)log(3x+1)−31+constant
The answer is:
−x+3(3x+1)log(3x+1)−31+constant
The answer (Indefinite)
[src]
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| 1 (3*x + 1)*log(3*x + 1)
| log(3*x + 1) dx = - - + C - x + ----------------------
| 3 3
/
3(3x+1)log(3x+1)−3x−1
The graph
34log4−3
=
−1+34log(4)
Use the examples entering the upper and lower limits of integration.