Solve $\int (4x+2)\sqrt{x^2+x+1}\,dx$

Question

Trying to solve this for a while now, so far I was able to come up without a proper answer.

Problem : $\displaystyle \int (4x+2)\sqrt{x^2+x+1}\,dx$.

I tried to take two common from $(4x+2)$ and also to take $(x+1)^2 - x$ from the root, but wasn't able to come up with something to take for substitution. A hint in the right direction would be highly appreciated

Edit :

I forgot to mention this. As this is part of the integration by substitution exercise it'd be highly appreciated if you could provide the hint in that direction.

Answer 1

Hint :

Are you asking $$ \int (4x+2)\sqrt{x^2+x+1}\ dx = \ldots? $$ If so, let $u=x^2+x+1\ \Rightarrow\ du=2x+1$.

Answer 2

I'm extremely new to integration and my methods are sloppy but I'd like to try out this question. The following isn't an answer but a demonstration that any substitution can do the job as long as it is done right (Although, a fundamental zen of calculus dictates that the rightest way is the fastest way, I think any path is better than no path)

I'm going to try my luck with $u = 4x+2 \implies \frac{du}{dx} = 4 \implies dx= \frac{1}{4} du$

Also, by the above assumption, $x = \frac{u-2}{4}$ $$ \require{cancel} \begin{align} &\int (4x+2)\sqrt{x^2+x+1}\,dx \\ &= \int{u\cdot\sqrt{\left(\frac{u-2}{4}\right)^2 + \frac{u-2}{4}+ 1}}\cdot\frac{1}{4} du\\ &= \frac{1}{4} \int{u}\cdot\sqrt{\frac{u^2 - 4u + 4 +4(u-2) + 4^2}{4^2}}\, du\\ &= \frac{1}{4} \int{ u \cdot \frac{1}{4}\cdot \sqrt{u^2 \cancel{- 4u} + 4 \cancel{+ 4u} - 8 +16}}\, du\\ &=\frac{1}{16} \int{u\cdot\sqrt{u^2 + 12}}\, du \end{align} $$ Mhh, we need to substitute again inorder to proceed. Let's continue this with $$v =\sqrt{ u^2 + 12 } \implies \frac{dv}{du} = \frac{2u}{2\sqrt{u^2 + 12}} = \frac{u}{v} \implies du = \frac{v}{u}\, dv$$ Continuing, $$ \begin{align} &\frac{1}{16} \int{u\cdot\sqrt{u^2 + 12}}\, du \\ &= \frac{1}{16} \int{u\cdot v \cdot \frac{v}{u}\, dv} \\ &= \frac{1}{16}\int{v^2}\, dv\\ &= \frac{1}{16}\frac{v^{2+1}}{(2+1)} + C\\ &= \frac{v^3}{48} + C \end{align}$$

Putting $v$ interms of $x$, you'll probably get the right answer.

$\dots$ Maybe taking $u=x^2+x+1$ would have been a better substitution. Well, practice makes perfect! :D

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{\root{x^{2} + x + 1} = x + t}$ we'll get $\ds{x = {1 - t^{2} \over 2t - 1}}$ such that

\begin{align} &\color{#c00000}{\int\pars{4x + 2}\root{x^{2} + x + 1}\,\dd x} =\int{32 t^{3} + 12t^{2} - 4 \over \pars{2t - 1}^{4}}\,\dd t \end{align}

Now, we set $\ds{t \equiv {1 - a \over 2}}$. Then

\begin{align} &\color{#c00000}{\int\pars{4x + 2}\root{x^{2} + x + 1}\,\dd x} =\int\pars{{27 \over 16 a^{4}} - {a^{2} \over 16} + {9 \over 16a^{2}} -{3 \over 16}}\,\dd a \\[5mm]&=-\frac{a^3}{48}-\frac{9}{16 a^3}-\frac{3 a}{16}-\frac{9}{16 a} \\[5mm]&=-\frac{1}{48} (1-2 t)^3-\frac{3}{16} (1-2 t)-\frac{9}{16 (1-2 t)} -\frac{9}{16 (1-2 t)^3} \\[5mm]&=-\frac{1}{48} \left[1-2 \left(\sqrt{x^2+x+1}-x\right)\right]^3-\frac{3}{16} \left[1-2 \left(\sqrt{x^2+x+1}-x\right)\right]-\frac{9}{16 \left[1-2 \left(\sqrt{x^2+x+1}-x\right)\right]}-\frac{9}{16 \left[1-2 \left(\sqrt{x^2+x+1}-x\right)\right]^3}\\[5mm]& + \mbox{a constant} \end{align}