Integral of x^2ln2xdx dx
The solution
Detail solution
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There are multiple ways to do this integral.
Method #1
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Rewrite the integrand:
x2log(2x)=x2log(x)+x2log(2)
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Integrate term-by-term:
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Let u=log(x).
Then let du=xdx and substitute du:
∫ue3udu
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Use integration by parts:
∫udv=uv−∫vdu
Let u(u)=u and let dv(u)=e3u.
Then du(u)=1.
To find v(u):
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Let u=3u.
Then let du=3du and substitute 3du:
∫3eudu
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The integral of a constant times a function is the constant times the integral of the function:
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The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 3eu
Now substitute u back in:
3e3u
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:
∫3e3udu=3∫e3udu
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Let u=3u.
Then let du=3du and substitute 3du:
∫3eudu
-
The integral of a constant times a function is the constant times the integral of the function:
-
The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 3eu
Now substitute u back in:
3e3u
So, the result is: 9e3u
Now substitute u back in:
3x3log(x)−9x3
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The integral of a constant times a function is the constant times the integral of the function:
∫x2log(2)dx=log(2)∫x2dx
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The integral of xn is n+1xn+1 when n=−1:
∫x2dx=3x3
So, the result is: 3x3log(2)
The result is: 3x3log(x)−9x3+3x3log(2)
Method #2
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Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=log(2x) and let dv(x)=x2.
Then du(x)=x1.
To find v(x):
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The integral of xn is n+1xn+1 when n=−1:
∫x2dx=3x3
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:
∫3x2dx=3∫x2dx
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The integral of xn is n+1xn+1 when n=−1:
∫x2dx=3x3
So, the result is: 9x3
Method #3
-
Rewrite the integrand:
x2log(2x)=x2log(x)+x2log(2)
-
Integrate term-by-term:
-
Let u=log(x).
Then let du=xdx and substitute du:
∫ue3udu
-
Use integration by parts:
∫udv=uv−∫vdu
Let u(u)=u and let dv(u)=e3u.
Then du(u)=1.
To find v(u):
-
Let u=3u.
Then let du=3du and substitute 3du:
∫3eudu
-
The integral of a constant times a function is the constant times the integral of the function:
-
The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 3eu
Now substitute u back in:
3e3u
Now evaluate the sub-integral.
-
The integral of a constant times a function is the constant times the integral of the function:
∫3e3udu=3∫e3udu
-
Let u=3u.
Then let du=3du and substitute 3du:
∫3eudu
-
The integral of a constant times a function is the constant times the integral of the function:
-
The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 3eu
Now substitute u back in:
3e3u
So, the result is: 9e3u
Now substitute u back in:
3x3log(x)−9x3
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The integral of a constant times a function is the constant times the integral of the function:
∫x2log(2)dx=log(2)∫x2dx
-
The integral of xn is n+1xn+1 when n=−1:
∫x2dx=3x3
So, the result is: 3x3log(2)
The result is: 3x3log(x)−9x3+3x3log(2)
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Now simplify:
9x3(3log(x)−1+log(8))
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Add the constant of integration:
9x3(3log(x)−1+log(8))+constant
The answer is:
9x3(3log(x)−1+log(8))+constant
The answer (Indefinite)
[src]
/
| 3 3 3
| 2 x x *log(2) x *log(x)
| x *log(2*x) dx = C - -- + --------- + ---------
| 9 3 3
/
∫x2log(2x)dx=C+3x3log(x)−9x3+3x3log(2)
The graph
3 3
1 log(2) e e *log(2*E)
- - ------ - -- + -----------
9 3 9 3
−9e3−3log(2)+91+3e3log(2e)
=
3 3
1 log(2) e e *log(2*E)
- - ------ - -- + -----------
9 3 9 3
−9e3−3log(2)+91+3e3log(2e)
1/9 - log(2)/3 - exp(3)/9 + exp(3)*log(2*E)/3
Use the examples entering the upper and lower limits of integration.