Integral of lnx^2 dx
The solution
Detail solution
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Let u=log(x).
Then let du=xdx and substitute du:
∫u2eudu
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Use integration by parts:
∫udv=uv−∫vdu
Let u(u)=u2 and let dv(u)=eu.
Then du(u)=2u.
To find v(u):
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The integral of the exponential function is itself.
∫eudu=eu
Now evaluate the sub-integral.
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Use integration by parts:
∫udv=uv−∫vdu
Let u(u)=2u and let dv(u)=eu.
Then du(u)=2.
To find v(u):
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The integral of the exponential function is itself.
∫eudu=eu
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:
∫2eudu=2∫eudu
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The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 2eu
Now substitute u back in:
xlog(x)2−2xlog(x)+2x
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Now simplify:
x(log(x)2−2log(x)+2)
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Add the constant of integration:
x(log(x)2−2log(x)+2)+constant
The answer is:
x(log(x)2−2log(x)+2)+constant
The answer (Indefinite)
[src]
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| 2 2
| log (x) dx = C + 2*x + x*log (x) - 2*x*log(x)
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∫log(x)2dx=C+xlog(x)2−2xlog(x)+2x
Use the examples entering the upper and lower limits of integration.