Integral of (sqrt(1-lnx))/x 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=1−log(x).
Then let du=−xdx and substitute −du:
∫(−u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫udu=−∫udu
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The integral of un is n+1un+1 when n=−1:
∫udu=32u23
So, the result is: −32u23
Now substitute u back in:
−32(1−log(x))23
Method #2
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Let u=x1.
Then let du=−x2dx and substitute −du:
∫−u1−log(u1)du
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The integral of a constant times a function is the constant times the integral of the function:
∫u1−log(u1)du=−∫u1−log(u1)du
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Let u=1−log(u1).
Then let du=udu and substitute du:
∫udu
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The integral of un is n+1un+1 when n=−1:
∫udu=32u23
Now substitute u back in:
32(1−log(u1))23
So, the result is: −32(1−log(u1))23
Now substitute u back in:
−32(1−log(x))23
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Add the constant of integration:
−32(1−log(x))23+constant
The answer is:
−32(1−log(x))23+constant
The answer (Indefinite)
[src]
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| ____________ 3/2
| \/ 1 - log(x) 2*(1 - log(x))
| -------------- dx = C - -----------------
| x 3
|
/
∫x1−log(x)dx=C−32(1−log(x))23
Use the examples entering the upper and lower limits of integration.