Integral of log2x 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=2x.
Then let du=2dx and substitute 2du:
∫4log(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫2log(u)du=2∫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: 2ulog(u)−2u
Now substitute u back in:
xlog(2x)−x
Method #2
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Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=log(2x) and let dv(x)=1.
Then du(x)=x1.
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 is the constant times the variable of integration:
∫1dx=x
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Now simplify:
x(log(2x)−1)
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Add the constant of integration:
x(log(2x)−1)+constant
The answer is:
x(log(2x)−1)+constant
The answer (Indefinite)
[src]
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| log(2*x) dx = C - x + x*log(2*x)
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22xlog(2x)−2x
22log2−2
=
−1+log(2)
Use the examples entering the upper and lower limits of integration.