Integral of dx/(x(loge(x))^1/3) 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:
x3log(e1)log(x)1=x3log(x)1
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There are multiple ways to do this integral.
Method #1
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Let u=x1.
Then let du=−x2dx and substitute −du:
∫−u3log(u1)1du
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The integral of a constant times a function is the constant times the integral of the function:
∫u3log(u1)1du=−∫u3log(u1)1du
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Let u=log(u1).
Then let du=−udu and substitute −du:
∫(−3u1)du
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The integral of a constant times a function is the constant times the integral of the function:
∫3u1du=−∫3u1du
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The integral of un is n+1un+1 when n=−1:
∫3u1du=23u32
So, the result is: −23u32
Now substitute u back in:
−23log(u1)32
So, the result is: 23log(u1)32
Now substitute u back in:
23log(x)32
Method #2
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Let u=log(x).
Then let du=xdx and substitute du:
∫3u1du
-
The integral of un is n+1un+1 when n=−1:
∫3u1du=23u32
Now substitute u back in:
23log(x)32
Method #2
-
Rewrite the integrand:
x3log(e1)log(x)1=x3log(x)1
-
Let u=x1.
Then let du=−x2dx and substitute −du:
∫−u3log(u1)1du
-
The integral of a constant times a function is the constant times the integral of the function:
∫u3log(u1)1du=−∫u3log(u1)1du
-
Let u=log(u1).
Then let du=−udu and substitute −du:
∫(−3u1)du
-
The integral of a constant times a function is the constant times the integral of the function:
∫3u1du=−∫3u1du
-
The integral of un is n+1un+1 when n=−1:
∫3u1du=23u32
So, the result is: −23u32
Now substitute u back in:
−23log(u1)32
So, the result is: 23log(u1)32
Now substitute u back in:
23log(x)32
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Add the constant of integration:
23log(x)32+constant
The answer is:
23log(x)32+constant
The answer (Indefinite)
[src]
/
| 2/3
| 1 3*log (x)
| ---------------- dx = C + -----------
| _________ 2
| / log(x)
| x* / -------
| 3 / / 1\
| \/ log\e /
|
/
∫x3log(e1)log(x)1dx=C+23log(x)32
The graph
Use the examples entering the upper and lower limits of integration.