how can find int ( 1 / (1+log x), int ( 2^x / (2+x) )
- Thread starter KKK_123
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Re: NEED HELP !
\(\displaystyle \int \frac{1}{1+ln(x)}dx\)
We can note that \(\displaystyle ln(e)=1\) and rewrite as \(\displaystyle \int \frac{1}{ln(e)+ln(x)}dx=\int \frac{1}{ln(ex)}dx\)
Now, we can make a sub of \(\displaystyle u=ex, \;\ x=\frac{u}{e}, \;\ dx=\frac{1}{e}du\)
\(\displaystyle \frac{1}{e}\int\frac{1}{ln(u)}du\)
Now, this is not possible in closed form. It is known as the logarithmic integral, notated as Li(x).
If it had limits, we may be able to do something, but an indefinite integral is not easily done.
Same for the second problem.
\(\displaystyle \int \frac{1}{1+ln(x)}dx\)
We can note that \(\displaystyle ln(e)=1\) and rewrite as \(\displaystyle \int \frac{1}{ln(e)+ln(x)}dx=\int \frac{1}{ln(ex)}dx\)
Now, we can make a sub of \(\displaystyle u=ex, \;\ x=\frac{u}{e}, \;\ dx=\frac{1}{e}du\)
\(\displaystyle \frac{1}{e}\int\frac{1}{ln(u)}du\)
Now, this is not possible in closed form. It is known as the logarithmic integral, notated as Li(x).
If it had limits, we may be able to do something, but an indefinite integral is not easily done.
Same for the second problem.
No, not in closed form. That is why they have given these types of integrals their own names.
\(\displaystyle \int_{2}^{x}\frac{1}{ln(u)}du=Li(x)(\text{logarithmic integral})\)
Like I said, if we had limits to work with we may be able to do something fancy and evaluate, but as is in indefinite form...no.
I ran yours through Maple and it gave me:
\(\displaystyle \int\frac{1}{1+ln(x)}dx=-\frac{1}{e}Ei(1,-ln(x)-1)\)
If we had limits of integration, we could most likely do it.
For instance, \(\displaystyle \int_{2}^{\infty}ln^{n}(u)du\). We can should this is divergent only if n>1. In your case, n=-1, so divergent.
The Ei(x) stands for the Exponential Integral, \(\displaystyle Ei(x)=\int_{x}^{\infty}\frac{e^{-u}}{u}du\).
As for \(\displaystyle \int\frac{2^{x}}{2+x}dx\)
It gives \(\displaystyle \frac{-1}{4}Ei(1,-ln(2)x-2ln(2))\)
Again, in terms of the Exponential Integral.
These are beyond a normal calc class where one learns substitution, parts, and what not.
You could write them in terms of their Taylor series and integrate that. Just a thought.
\(\displaystyle \int_{2}^{x}\frac{1}{ln(u)}du=Li(x)(\text{logarithmic integral})\)
Like I said, if we had limits to work with we may be able to do something fancy and evaluate, but as is in indefinite form...no.
I ran yours through Maple and it gave me:
\(\displaystyle \int\frac{1}{1+ln(x)}dx=-\frac{1}{e}Ei(1,-ln(x)-1)\)
If we had limits of integration, we could most likely do it.
For instance, \(\displaystyle \int_{2}^{\infty}ln^{n}(u)du\). We can should this is divergent only if n>1. In your case, n=-1, so divergent.
The Ei(x) stands for the Exponential Integral, \(\displaystyle Ei(x)=\int_{x}^{\infty}\frac{e^{-u}}{u}du\).
As for \(\displaystyle \int\frac{2^{x}}{2+x}dx\)
It gives \(\displaystyle \frac{-1}{4}Ei(1,-ln(2)x-2ln(2))\)
Again, in terms of the Exponential Integral.
These are beyond a normal calc class where one learns substitution, parts, and what not.
You could write them in terms of their Taylor series and integrate that. Just a thought.