Integral of ln(ln(x)) dx
The solution
Detail solution
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Let u=log(x).
Then let du=xdx and substitute du:
∫eulog(u)du
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There are multiple ways to do this integral.
Method #1
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Let u=log(u).
Then let du=udu and substitute du:
∫ueueeudu
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Use integration by parts:
∫udv=uv−∫vdu
Let u(u)=u and let dv(u)=eueeu.
Then du(u)=1.
To find v(u):
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Let u=eu.
Then let du=eudu and substitute du:
∫eudu
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The integral of the exponential function is itself.
∫eudu=eu
Now substitute u back in:
Now evaluate the sub-integral.
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Let u=eu.
Then let du=eudu and substitute du:
∫ueudu
EiRule(a=1, b=0, context=exp(_u)/_u, symbol=_u)
Now substitute u back in:
Ei(eu)
Now substitute u back in:
eulog(u)−Ei(u)
Method #2
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Use integration by parts:
∫udv=uv−∫vdu
Let u(u)=log(u) and let dv(u)=eu.
Then du(u)=u1.
To find v(u):
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The integral of the exponential function is itself.
∫eudu=eu
Now evaluate the sub-integral.
EiRule(a=1, b=0, context=exp(_u)/_u, symbol=_u)
Now substitute u back in:
xlog(log(x))−Ei(log(x))
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Add the constant of integration:
xlog(log(x))−Ei(log(x))+constant
The answer is:
xlog(log(x))−Ei(log(x))+constant
The answer (Indefinite)
[src]
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| log(log(x)) dx = C - Ei(log(x)) + x*log(log(x))
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∫log(log(x))dx=C+xlog(log(x))−Ei(log(x))
The graph
(-0.577215664901533 + 3.14159265358979j)
(-0.577215664901533 + 3.14159265358979j)
Use the examples entering the upper and lower limits of integration.