How to present this particular integration

Question

$$ \int\frac{x^2+3x}{\sqrt{x^2+6x+10}}dx $$ How to present this integral?

Answer 1

Hint:

Use the substitution $\; x+3=\sinh t,\enspace \mathrm d\mkern1mu x=\cosh t\,\mathrm d\mkern1mu t$.

Answer 2

HINT:

$$\int\frac{x^2+3x}{\sqrt{x^2+6x+10}}\space\text{d}x=\int\frac{x^2+3x}{\sqrt{(x+3)^2+1}}\space\text{d}x=$$

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Substitute $u=x+3$ and $\text{d}u=\text{d}x$:

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$$\int\frac{(u-3)^2+3(u-3)}{\sqrt{u^2+1}}\space\text{d}u=$$

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Substitute $u=\tan(s)$ and $\text{d}u=\sec^2(s)\text{d}s$.

Then $\sqrt{u^2+1}=\sqrt{\tan^2(s)+1}=\sec(s)$ and $s=\arctan(u)$:

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$$\int\left(\left(\tan(s)-3\right)^2+3\left(\tan(s)-3\right)\right)\sec(s)\space\text{d}s=$$ $$\int\tan^2(s)\sec(s)\space\text{d}s-3\int\tan(s)\sec(s)\space\text{d}s=$$

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Use $\tan^2(s)=\sec^2(s)-1$:

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$$\int\sec^3(s)\space\text{d}s-\int\sec(s)\space\text{d}s-3\int\tan(s)\sec(s)\space\text{d}s=$$

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Now, use the reduction formula:

$$\int\sec^m(s)\space\text{d}s=\frac{\sin(s)\sec^{m-1}(s)}{m-1}+\frac{m-2}{m-1}\int\sec^{m-2}(s)\space\text{d}s$$

And for $\int\tan(s)\sec(s)\space\text{d}s$, substitute $q=\sec(s)$ and $\text{d}q=\tan(s)\text{d}s$:

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$$\int\sec^3(s)\space\text{d}s-\int\sec(s)\space\text{d}s-3\int1\space\text{d}q=$$ $$\int\sec^3(s)\space\text{d}s-\int\sec(s)\space\text{d}s-3q=$$

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Use:

$$\int\sec(s)\space\text{d}s=\ln\left|\sec(s)+\tan(s)\right|+\text{C}$$

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$$\int\sec^3(s)\space\text{d}s-\ln\left|\sec(s)+\tan(s)\right|-3\sec(s)$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With the Euler sub$\ds{\ldots\root{x^{2} + 6x + 10} - x \equiv t\ \imp\ x = -\,\half\,{t^{2} - 10 \over t - 3}}$ the integral is reduced to: \begin{align} \int{x^{2} + 3x \over \root{x^{2} + 6x + 10}}\,\dd x & = -\,{1 \over 4}\int{\pars{t^{2} - 10}\pars{t - 2 }\pars{t - 4} \over \pars{t - 3}^{3}}\,\dd t \end{align} which can be integrated by partial fraction expansion.

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For instance, $$ \left\lbrace\begin{array}{rcl} \ds{t^{2} - 10} & \ds{=} & \ds{\pars{t - 3}^{2} + 6\pars{t - 3} - 1} \\[2mm] \ds{\pars{t - 2}\pars{t - 4}} & \ds{=} & \ds{\pars{t - 3}^{2} - 1} \end{array}\right. $$ such that \begin{align} \pars{t^{2} - 10}\pars{t - 2 }\pars{t - 4} = \pars{t - 3}^{4} + 6\pars{t - 3}^{3} - 2\pars{t - 3}^{2} - 6\pars{t - 3} + 1 \end{align}