Integral of 4/(x(ln^2x+1)) dx
The solution
Detail solution
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The integral of a constant times a function is the constant times the integral of the function:
∫x(log(x)2+1)4dx=4∫x(log(x)2+1)1dx
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Don't know the steps in finding this integral.
But the integral is
RootSum(4z2+1,(i↦ilog(2i+log(x))))
So, the result is: 4RootSum(4z2+1,(i↦ilog(2i+log(x))))
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Now simplify:
−2ilog(log(x)−i)+2ilog(log(x)+i)
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Add the constant of integration:
−2ilog(log(x)−i)+2ilog(log(x)+i)+constant
The answer is:
−2ilog(log(x)−i)+2ilog(log(x)+i)+constant
The answer (Indefinite)
[src]
/
|
| 4 / 2 \
| --------------- dx = C + 4*RootSum\4*z + 1, i -> i*log(2*i + log(x))/
| / 2 \
| x*\log (x) + 1/
|
/
∫x(log(x)2+1)4dx=C+4RootSum(4z2+1,(i↦ilog(2i+log(x))))
The graph
oo
/
|
| 1
4* | --------------- dx
| / 2 \
| x*\1 + log (x)/
|
/
1
41∫∞x(log(x)2+1)1dx
=
oo
/
|
| 1
4* | --------------- dx
| / 2 \
| x*\1 + log (x)/
|
/
1
41∫∞x(log(x)2+1)1dx
4*Integral(1/(x*(1 + log(x)^2)), (x, 1, oo))
Use the examples entering the upper and lower limits of integration.