# Integrating $\int\frac{-3}{x^2+4}\ dx$

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.

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

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 $$\int\dfrac {-3}{x^2 + 4}\,\mathrm dx = -\dfrac32\arctan\left(\dfrac x2\right)+C.$$

• 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 $-\int\dfrac 6{4t^2 + 4}\,\mathrm dt$
– Burt
Dec 3, 2019 at 3:35
• Substitute $x=2t$ Dec 3, 2019 at 3:40
• @J.W.Tanner what do you mean?
– Burt
Dec 3, 2019 at 3:47
• 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 Dec 3, 2019 at 3:56
• @Burt Which part are you stuck at?
– an4s
Dec 3, 2019 at 5:00

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?

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

-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）