# Evaluating the primitive $\int \frac{\mathrm dx}{e^{2x} + e^x + 1}$

Could someone help me evaluate this? $$\int \frac{\mathrm dx}{e^{2x} + e^x + 1}$$ I tried to solve it for hours with no success. I tried Wolframalpha but it's giving a step by step solution that is too long that in an exam I won't even have the time to write the solution.

• Please refrain from making non-inline math in titles. – user93957 Jan 4 '14 at 21:19

Let $u = e^x$. Then, $du = e^x\,dx$, or, equivalently, $dx = \frac{1}{u}du$. Thus, the integral becomes: \begin{align} \int\frac{dx}{e^{2x}+e^x+1} &= \int\frac{du}{(u^2+u+1)u}\\ &= \int\frac{du}{(u^2+u+1)u} \end{align} Now, hit it with partial fractions: \begin{align} \frac{1}{(u^2+u+1)u} &= \frac{Au+B}{u^2+u+1} + \frac{D}{u}\\ &=\frac{Au^2+Bu+Du^2+Du+D}{(u^2+u+1)u}\\ &=\frac{(A+D)u^2+(B+D)u+D}{(u^2+u+1)u} \end{align} Thus: $$A+D = 0\\ B+D = 0\\ D = 1$$

So, the integral is: $$\int\frac{du}{(u^2+u+1)u} = \underbrace{\int \frac{-u-1}{u^2+u+1} du}_{\text{integral 1}} + \underbrace{\int \frac{1}{u}du}_{\text{integral 2}}$$

Integral $2$ is trivial, so I won't write it out. For integral $1$, we apply the substitution $w = u^2+u+1$, so $dw = (2u + 1)du$. \begin{align} \int \frac{-u-1}{u^2+u+1} du &= \frac{-1}{2}\int \frac{2u+1}{u^2+u+1}du + \frac{-1}{2}\int\frac{1}{u^2+u+1}du\\ &= \frac{-1}{2}\int \frac{1}{w}dw - \frac{1}{2} \int \frac{1}{u^2 + u+1}du\\ &= \frac{-1}{2}\ln|w| - \frac{1}{2} \underbrace{\int \frac{1}{u^2 + u+1}}_{\text{integral 3}} \end{align}

For Integral $3$, we use our knowledge of the derivative of arctangent. We have: \begin{align} \int \frac{1}{u^2 + u+1}du &= \int \frac{1}{\left(u+\frac{1}{2}\right)^2 + \frac{3}{4}}du\\ &= \int \frac{1}{\left(\frac{u+\frac{1}{2}}{\sqrt{3}/2}\right)^2 + 1}du\\ &= \int \frac{1}{\left(\frac{2u+1}{\sqrt{3}}\right)^2 + 1}du\\ &= \frac{2}{\sqrt{3}}\arctan\left(\frac{2u+1}{\sqrt{3}}\right) +C \end{align}

Thus, our overall answer is: \begin{align} \int\frac{dx}{e^{2x}+e^x+1} &= I_1 + \ln|e^x| \\ &= \frac{-1}{2}\ln|e^{2x}+e^x+1| - \frac{1}{\sqrt{3}}\arctan\left(\frac{2e^x+1}{\sqrt{3}}\right) + x + C \end{align}

• Apparently, I'm a bit late to the party here... :) – apnorton Jan 4 '14 at 21:50
• I am accepting this as the answer since you actually took the trouble of writing a complete step by step solution :D. I had already solved it though. – Ahmed Ali Jan 4 '14 at 21:53

Substitute $u=e^x,$ then use partial fractions.

Hint. You can calculate the integral of any rational function of $e^x$ by making the substitution $u = e^x$.

• My answer is too similar to Igor Rivin's. How do I remove my answer. – user119107 Jan 4 '14 at 21:25
• There is a button for that. – user93957 Jan 4 '14 at 21:26
• Where? All I see is "share edit flag". – user119107 Jan 4 '14 at 21:32
• That's because you aren't registered yet, I guess. – user93957 Jan 4 '14 at 21:33
• It's strange that I'm required to have "privileges" to remove my own answer. – user119107 Jan 4 '14 at 21:42

\begin{align} u & = e^x \\ du & = e^x\,dx = u\,dx \\ du/u & = dx \end{align} $$\int \frac{dx}{e^{2x}+e^x+1} = \int\frac{du/u}{u^2+u+1}$$ \begin{align} u^2+u+1 & = \left(u^2 + u + \frac14\right) + \frac34\quad \text{(completing the square)} \\[10pt] & = \left(u+\frac12\right)^2+\frac34 \\[10pt] & = \frac34\left(\frac43\left(u+\frac12\right)^2 + 1\right) \\[10pt] & = \frac32\left(w^2+1\right) \text{ where } w = \frac{2}{\sqrt{3}}\left(u+\frac12\right) = \frac{1}{\sqrt{3}} (2u+1) \\[10pt] & {}\qquad\qquad\qquad\text{ and }u = \frac{\sqrt{3}\ w - 1}{2}. \end{align} So the integral is $$\int\frac{2\,dw}{(\sqrt{3}\ w-1)(w^2+1)} = \int \left(\frac{A\,dw}{\sqrt{3}\ w+1} + \frac{B\,dw}{w^2+1} \right).$$ So you get a logarithm and an arctangent. Of course, you have to do a bit of algebra to figure out what $A$ and $B$ are.

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \color{#00f}{\large\int{\dd x \over \expo{2x} + \expo{x} + 1}}&= -\int{\overbrace{\expo{-x}}^{\ds{\equiv\ t}}\pars{-\expo{-x}\,\dd x} \over 1 + \expo{-x} + \expo{-2x}} =-\int{t\,\dd t \over t^{2} + t + 1} =-\int{t\,\dd t \over \pars{\overbrace{t + 1/2}^{\ds{\equiv\ y}}}^{2} + 3/4} \\[3mm]&=-\int{y\,\dd y \over y^{2} + 3/4} + \half\int{\dd t \over y^{2} + \pars{\root{3}/2}^{2}} \\[3mm]&=-\,\half\,\ln\pars{y^{2} + {3 \over 4}} + {\root{3} \over 3}\,\arctan\pars{{2\root{3} \over 3}\,y} \\[3mm]&=-\,\half\,\ln\pars{\bracks{t + \half}^{2} + {3 \over 4}} + {\root{3} \over 3}\,\arctan\pars{{2\root{3} \over 3}\,\bracks{t + \half}} \\[3mm]&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!=\color{#00f}{\large -\,\half\,\ln\pars{\expo{-2x} + \expo{-x} + 1} + {\root{3} \over 3}\,\arctan\pars{{2\root{3} \over 3}\,\bracks{\expo{-x} + \half}}} \\[3mm]&+\ \mbox{a constant} \end{align}