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Find$$\displaystyle \int \dfrac{6x+4}{x^2+4}dx$$

I'm not really sure where to begin with this one - I know the answer will probably involve an $\arctan$, but I am unsure on how to use $\arctan$ in integrating. A full step by step explanation would be really appreciated.

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    $\begingroup$ $$6x+4=3\dfrac{d(x^2+4)}{dx}+4$$ $$\int\frac{dx}{x^2+2^2}=?$$ $\endgroup$ – lab bhattacharjee Sep 2 '14 at 18:43
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HINT

Decompose the integral as

$$\int \frac{6x+4}{x^2+4}~\mathrm{d}x = 3\int \frac{2x}{x^2+4}~\mathrm{d}x+4\int\frac{1}{x^2+4}~\mathrm{d}x$$

The first integral on the right will involve $\ln$ and the second will involve $\arctan$.

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  • $\begingroup$ Thanks for your post, I got the first integral, but I'm unsure about the second, particularly how arctan works. I've tried googling it but I don't really understand the explanations I found. $\endgroup$ – Jim Sep 2 '14 at 18:57
  • $\begingroup$ @Jim Hint: let $x=2\tan(\theta).$ $\endgroup$ – beep-boop Sep 2 '14 at 19:00
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    $\begingroup$ @jim In general, $$\int \frac{1}{a^2+x^2} \mathrm{d}x=\frac{1}{a}\arctan\left(\frac{x}{a}\right).$$ $\endgroup$ – beep-boop Sep 2 '14 at 19:01
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    $\begingroup$ Hopefully you know that $\arctan x$ differentiates to give $\displaystyle{\frac{1}{x^2+1}}$. It's all about trying to make the second integral look like this. Why not divide numerator and denominator by $4$: $$\frac{1}{x^2+4} \equiv \frac{1/4}{x^2/4+1} \equiv \frac{1}{4}\left(\frac{1}{\left(\frac{x}{2}\right)^{\! 2}+1}\right)$$ Then you can make the substitution $u=\frac{x}{2}$. $\endgroup$ – Fly by Night Sep 2 '14 at 19:01
  • $\begingroup$ Dammit. Forgot my $+C$ at the end, and my five minutes of editing time are over! $\endgroup$ – beep-boop Sep 2 '14 at 19:08
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First, you can split the integral up as follows (this is what lab bhattacharjee is referring to):

$$\int\frac{6x+4}{x^2+4}dx=3\int\frac{2x}{x^2+4}dx+4\int\frac{1}{x^2+4}dx.$$

The motivation for doing this is two-fold:

  1. To isolate the function which integrates to something like $\arctan$ - you recognised this.

  2. To obtain an integrand whose top is the derivative of the bottom - this involves $\ln$.

It is a "standard fact" (which means that it is useful to rememeber) that $$\int \frac{1}{1+x^2}dx=\arctan(x) + C.$$ You can prove this by using the substitution $x=\tan \theta$.

Now, you are required to find $$\int\frac{1}{x^2+4}dx$$ What happens if you make the substitution $x=2u$?

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  • $\begingroup$ I think you're missing a constant of integration: $$\int \frac{1}{1+x^2}~\mathrm{d}x = \arctan x + C$$ $\endgroup$ – Fly by Night Sep 2 '14 at 19:07
  • $\begingroup$ I'm not sure I understand. The OP wants an anti-derivative. What is the connection between limits and Riemann integration (to find areas), and finding anti-derivatives? There are many functions that are not Riemann integrable on the entire real line that have well-defined anti-derivatives. $\endgroup$ – Fly by Night Sep 2 '14 at 19:16
  • $\begingroup$ You are correct, I suppose the OP wanted an antiderivative (although, to be pedantic, when a constant of integration is introduced, you are providing the entire family of antiderivatives). Indeed there exist non-Riemann integrable functions with an antiderivative and conversely there are Riemann integrable functions which do not have an antiderivative. $\endgroup$ – Elliptic2005 Sep 2 '14 at 19:19
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$$ x^2 + 4 = \frac{1}{4}\left( {\left( {\frac{x}{2}} \right)^2 + 1} \right) \Rightarrow \int {\frac{{dx}}{{x^2 + 4}}} = 4\int \frac{{dx}}{{\left( {\frac{x}{2}} \right)^2 + 1}} $$

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