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I've been trying this one for days and I can't seem to get it. Any ideas?

$\int \frac{6x-15}{x^2 +8x+101}dx$

Thanks, Nick

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  • $\begingroup$ Use $x + 4 \equiv t$. $\endgroup$ – Felix Marin Sep 2 '14 at 23:15
  • $\begingroup$ That's a variable change. It's easy to work with $t$. At the end you can go back to $x = t - 4$. See $\color{#00f}{\tt @alexqwx}$ answer below. $\endgroup$ – Felix Marin Sep 2 '14 at 23:30
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Hint: $$\Large \frac{6x-15}{x^2+8x+101} \equiv 3\left[\frac{2x+8}{x^2+8x+101}\right]-39 \left[\frac{1}{x^2+8x+101}\right].$$

The first part should be easy to integrate (inspection): the second requires a trigonometric substitution.

Hint 2: $$\frac{1}{x^2+8x+101} \equiv \frac{1}{\color{green}{(x+4)^2}+\color{red}{85}}.$$

Now let $x+4 \equiv \sqrt{85} \tan(\theta).$


If Hint 2 is too tedious (i.e. you can't be bothered going through the whole process of substitution), you might want to use the standard result $$\int \frac{1}{\color{green}{u^2}+\color{red}{a^2}}\ \mathrm{d}u =\frac{1}{a}\arctan\left(\frac{u}{a}\right)+C \quad.$$

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  • $\begingroup$ Thank you so much! I totally get it now. $\endgroup$ – Nick James Sep 2 '14 at 23:38
  • $\begingroup$ @NickJames Sure! The key here is that the integrand is "nearly" of the form $\frac{f'(x)}{f(x)}$. In order to get it into such a form, we have a bit left over, which is where the pesky $$-39 \left[\frac{1}{x^2+8x+101}\right]$$ comes from. $\endgroup$ – beep-boop Sep 2 '14 at 23:46
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Hint; Write the integral as $$\int \frac{3(2x+8)}{x^2+8x+101}-\frac{39}{x^2+8x+101} dx$$ For the first let $$u=x^2+8x+101$$ and for the second rewrite it as $$39 \int \frac{1}{(x+4)^2+85} dx$$ and let $$v=x+4$$

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$\displaystyle\int\dfrac{6x-15}{x^2+8x+101} dx=\int\dfrac{6x-15}{(x+4)^2+85} dx$.

Now let $u=x+4$, so $x=u-4$ and $dx=du$ to get

$\displaystyle\int\frac{6u-39}{u^2+85} du=3\int\frac{2u}{u^2+85} du-39\int\frac{1}{u^2+85} du$

$=3\ln(u^2+85)-39\left(\frac{1}{\sqrt{85}}\arctan\frac{u}{\sqrt{85}}\right)+C$.

(Now substitute back for u.)

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