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I was given the problem: $\int\frac{-3}{x^2+4}\ dx $

I am unsure how to integrate it. It seems to me that it requires a u-substitution because of the $x^2$ in the denominator, but I cannot figure out what to substitute. I cannot substitute in $x^2$ because then I am stuck with an x in that derivative.
I do know that I can pull the -3 out of the integral and get: $-3\int\frac{1}{x^2+4}$ but this does not help me - I'm still stuck. I also know that arctan is $\int\frac{1}{x^2+1}\ dx$ which seems similar to this, but it is not the same thing.

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  • $\begingroup$ Divide by $4$ and define $y = x/2$ then solve as an arctan $\endgroup$
    – fGDu94
    Dec 3, 2019 at 3:22
  • $\begingroup$ Let $x=2 \tan(\theta)$. $\endgroup$ Dec 3, 2019 at 3:24

4 Answers 4

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Substitute $t = \dfrac x2\implies x = 2t,\mathrm dx = 2\mathrm dt$.

Therefore,

$$\int\dfrac {-3}{x^2 + 4}\,\mathrm dx = -\int\dfrac 6{4t^2 + 4}\,\mathrm dt = -\int\frac6{4(t^2 + 1)}\,\mathrm dt = -\frac 32\int\frac1{t^2 + 1}\,\mathrm dt$$

$\displaystyle\int\dfrac 1{t^2 + 1}\,\mathrm dt$ results in $\arctan t + C$. Reverse substitution to get $$\int\dfrac {-3}{x^2 + 4}\,\mathrm dx = -\dfrac32\arctan\left(\dfrac x2\right)+C.$$

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  • $\begingroup$ I can't follow how that substitution works. Can you break it down really simply - between $\int\dfrac {-3}{x^2 + 4}\,\mathrm dx$ and $-\int\dfrac 6{4t^2 + 4}\,\mathrm dt$ $\endgroup$
    – Burt
    Dec 3, 2019 at 3:35
  • $\begingroup$ Substitute $x=2t$ $\endgroup$ Dec 3, 2019 at 3:40
  • $\begingroup$ @J.W.Tanner what do you mean? $\endgroup$
    – Burt
    Dec 3, 2019 at 3:47
  • $\begingroup$ If $x=2t$ then $\displaystyle\int\dfrac{-3}{x^2+4}dx=\int\dfrac{-3}{(2t)^2+4}2dt$; I thought that's what you were asking for $\endgroup$ Dec 3, 2019 at 3:56
  • $\begingroup$ @Burt Which part are you stuck at? $\endgroup$
    – an4s
    Dec 3, 2019 at 5:00
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Yup, $\tan^{-1}$ is a great start.

So once you have $\displaystyle -3\int \frac{1}{x^2+4}\,dx$, you can turn this into $\displaystyle -\frac{3}{4}\int\frac{1}{(\frac{x}{2})^2+1}\,dx=-\frac{3}{2}\int\frac{\frac{1}{2}}{(\frac{x}{2})^2+1}\,dx$.

You see the perfect u-substitution yet?

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$$\int{\frac{-3dx}{x^2+4}}=-3\int{\frac{dx}{x^2+4}}=-3\int{\frac{dx}{4\left(\frac{x^2}{4}+1\right)}}$$

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-3/2arctanx/2 is a primitive function of -3/(x^2+4). It is easy to check this because (-3/2arctanx/2)’=-3/(x^2+4)

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