# I need help calculating the anti derivative of $f(x)=2* \frac{c^{2x+1}}{2x+1}$

Recently I asked a question where as answer came out $\ln{(a)}=2*\text{arctanh } u$ which is to be calculated using $\sum_{k=0}^{\infty}(\frac{c^{2k+1}}{2k+1})$ with $u=\frac{a-1}{a+1} < 1$ if that helps.

Now since I need to approach and not necessarily calculate $\ln{a}$, I figured that if I used it like the Riemann-sums, the antiderivative of the function can also be used, since the sum is leading to an area of $1*f(x)$ and the antiderivative will approximately reach that value (not quite exact, but like I said, that was not necessary).

Anyways, my question was, what is the antiderivative of $f(x)=2* \frac{c^{2x+1}}{2x+1}$? I don't really know if it's very hard, I'm just very bad at calculating antiderivatives when they're not as simple :s

• Subbing $t:=2x+1$ and $\mathrm dt=2~\mathrm dx$, we have $\int\frac{c^t}t~\mathrm dt$. Further subbing $p:=\log t$ and $\mathrm dp=\frac{\mathrm dt}t$ gives $\int c^{e^p}~\mathrm dp=\operatorname{Ei}(e^x\log c)+C$ – Prasun Biswas Dec 26 '17 at 9:51
• Edit: Should be $\operatorname{Ei}(e^p\log c)$ since $e^{\log(c^{e^p})}=e^{e^p\log c}$. Simplifying should give the answer to be $\operatorname{Ei}((2x+1)\log c)+C$ with C, an arbitrary constant. – Prasun Biswas Dec 26 '17 at 10:00
• Using 2 artanh u = log ((1 + u)/(1 - u)) might be easier. – random Dec 26 '17 at 10:28

$$I = \int 2 \frac{c^{2x+1}}{2x+1}\, dx = \int \frac{c^u}{u}\, du$$ substituting $u = 2x+1$. Now, this is an exponential integral given by: $$I = \operatorname{Ei}\bigl((2x+1)\ln c\bigr) + C$$
• I think it should be $\operatorname{Ei}(e^x\ln c)+C$ – Prasun Biswas Dec 26 '17 at 9:53
• Oh right, I made a mistake computing $\operatorname{Ei}(e^p\ln c)=\operatorname{Ei}(e^{\ln(2x+1)}\ln c)=\operatorname{Ei}((2x+1)\ln c)$ – Prasun Biswas Dec 26 '17 at 9:57