Integral of ln(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=x+1.
Then let du=dx and substitute du:
∫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
Now substitute u back in:
−x+(x+1)log(x+1)−1
Method #2
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Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=log(x+1) and let dv(x)=1.
Then du(x)=x+11.
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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Rewrite the integrand:
x+1x=1−x+11
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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:
∫1dx=x
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The integral of a constant times a function is the constant times the integral of the function:
∫(−x+11)dx=−∫x+11dx
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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)
So, the result is: −log(x+1)
The result is: x−log(x+1)
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Now simplify:
−x+(x+1)log(x+1)−1
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Add the constant of integration:
−x+(x+1)log(x+1)−1+constant
The answer is:
−x+(x+1)log(x+1)−1+constant
The answer (Indefinite)
[src]
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| log(x + 1) dx = -1 + C - x + (x + 1)*log(x + 1)
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∫log(x+1)dx=C−x+(x+1)log(x+1)−1
The graph
−1+2log(2)
=
−1+2log(2)
Use the examples entering the upper and lower limits of integration.