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I am having trouble with this integral: $\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx.$ One obvious thing would be to complete the square: $x^2+2x+25=(x+1)^2+24$. But then, I don't know which substitution to use. Can anyone help? Thank you.

  • 1
    Try $u=x+1$ followed by a trig substitution.2012-05-06

4 Answers 4

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I believe the substitution you're looking for is

$x+1=2\sqrt6\tan\theta,x=2\sqrt6\tan\theta-1$.

This will reduce the denominator to

$\sqrt{(2\sqrt6\tan\theta)^2+24}=\sqrt{24\tan^2\theta+24}=\sqrt{24\sec^2\theta}=2\sqrt6\sec\theta$

I assume you'll be able to take it from there?

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    Thanks! This was easier than I thought!2012-05-07
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You will have, by substitution of $y=\frac{x+1}{\sqrt{24}}$,

$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx=\int_\frac{1}{\sqrt{24}}^1\frac{(\sqrt{24}y-1)}{\sqrt{y^2+1}}dy$

that becomes

$\int_\frac{1}{\sqrt{24}}^1\frac{(\sqrt{24}y-1)}{\sqrt{y^2+1}}dy=\sqrt{24}\int_\frac{1}{\sqrt{24}}^1\frac{y}{\sqrt{y^2+1}}dy-\int_\frac{1}{\sqrt{24}}^1\frac{1}{\sqrt{y^2+1}}dy.$

This gives finally

$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx=\sqrt{48} - \sqrt{26} + {\rm arcsinh}\left(\frac{1}{\sqrt{24}}\right) - {\rm arcsinh}(1).$

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    @Mike: Sorry for the mistake. I think now it should be better.2012-05-06
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We can do this in still another way (what I might refer to as the 'naive' way - set everything under the radical to a variable and go). You started with

$\int_0^4\frac{xdx}{\sqrt{x^2+2x+25}} = \int_0^4 \frac{xdx}{\sqrt{(x+1)^2 + 24}}$

Let $u = (x+1)^2 + 24$, so that $du = (2x+2)dx$

Then $\displaystyle \int_0^4 \frac{xdx}{\sqrt{(x+1)^2 + 24}} = \frac{1}{2}\int_0^4 \frac{2x + 2}{\sqrt{(x+1)^2 + 24}}dx - \int_0^4 \frac{dx}{\sqrt{(x+1)^2+ 24}}$

The second integral is a standard arcsinh integral, and with the change of variables above the first integral becomes $\displaystyle \frac{1}{2}\int \frac{du}{\sqrt u}$

I always like it when an integral can be computed in many ways.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With Euler First Substitution $\ds{x = {25 - t^{2} \over 2\pars{t - 1}}}$:

\begin{align} \int_{0}^{4}{x\,\dd x \over \root{x^{2} + 2x + 26}} & = \int_{3}^{5}\bracks{{12 \over \pars{t - 1}^{2}} - {1 \over t - 1} - {1 \over 2}}\,\dd t = 3 - \ln\pars{2} - 1 = \bbx{2 - \ln\pars{2}} \end{align}