# $\int \min \{\frac1{(1+x^2)^2},\frac1{5-x^2}\}$ solution verification

$$\int \min \{\frac1{(1+x^2)^2},\frac1{5-x^2}\}$$

I see this integral as the integral of the function which smaller of these two

$$\int\frac1{(1+x^2)^2}=\frac12\arctan(x)+\frac x{2(1+x^2)}+C_1$$

$$\int\frac1{5-x^2}=\frac{\sqrt5}{10}\log|\sqrt5+x| -\frac{\sqrt5}{10}\log|\sqrt5-x| +C_2$$

$$\int \min \{\frac1{(1+x^2)^2},\frac1{5-x^2}\}= \cases{ \frac12\arctan(x)+\frac x{2(1+x^2)}+C_1 & \frac1{(1+x^2)^2}< \frac1{5-x^2} \cr \frac{\sqrt5}{10}\log|\sqrt5+x| -\frac{\sqrt5}{10}\log|\sqrt5-x| +C_2 & \frac1{5-x^2}< \frac1{(1+x^2)^2} }$$

Is the idea correct and what happens in points where $$\frac1{(1+x^2)^2}=\frac1{5-x^2}$$ ?

• You are not done, the antiderivative should be continuous (except at asymptotes). – Yves Daoust Jun 15 '20 at 12:12
• So i need to check continuity in points where $\frac 1{(1+x^2)^2}=\frac 1{5-x^2}\$ – Milan Jun 15 '20 at 12:17
• Yes, that's it. – Yves Daoust Jun 15 '20 at 12:21
• @YvesDaoust What if it isn't continous ? Does then the antiderivative not exist ? – Milan Jun 15 '20 at 12:27

## 2 Answers

For $$x^2>5$$, the function reduces to $$\dfrac1{5-x^2}$$ and we can use an indefinite integral

$$F(x)=\int\frac{dx}{5-x^2}+C_0.$$ Notice that you may not cross the borders $$x^2=5$$ and there can be two distinct constants on either sides.

For $$x^2<5$$, we first restrict the study to $$x\ge0$$ for convenience. We switch from one function to the other at $$x=1$$. So for $$x\le1$$, we adopt $$F(x)=\int_0^x\frac{dt}{5-t^2}+C_1$$ and for $$1\le x\le \sqrt5,$$

$$F(x)=\int_0^1\frac{dt}{5-t^2}+\int_1^x\frac{dt}{(1+t^2)^2}+C_1=\frac1{\sqrt 5}\text{artanh}\frac1{\sqrt 5}+\int_1^x\frac{dt}{(1+t^2)^2}+C_1.$$

Now for the negative domain, we can use by symmetry

$$F(x)= C_1-F(-x).$$

What did you do to determine which values of x make one smaller than the other? First notice that $$\frac{1}{(1+ x^2)^2}$$ is always positive while $$\frac{1}{5- x^2}$$ is positive only for $$-\sqrt{5}< x< \sqrt{5}$$.

For all $$x< -\sqrt{5}$$ and all $$x> \sqrt{5}$$, the minimum if $$\frac{1}{5- x^2}$$. For $$-\sqrt{5}< x< \sqrt{5}$$, $$\frac{1}{(1+ x^2)^2}< \frac{1}{5- x^2}$$ if and only if $$5- x^2< (1- x^2)^2= 1- 2x^2+ x^4$$.

So we want to find x such that $$x^4- x^2- 4> 0$$. Let $$y= x^2$$ and solve $$y^2- y- 4> 0$$. $$y^2- y- 4= y^2- y+ \frac{1}{4}- \frac{1}{4}- 4= (y- \frac{1}{2})^2- \frac{17}{4}> 0$$. $$(y- \frac{1}{2})^2> \frac{17}{4}$$.

That will be true for $$y= x^2< \frac{1- \sqrt{17}}{2}$$ and for $$y= x^2> \frac{1+ \sqrt{17}}{2}$$. Of course, for real x, $$x^2$$ is not negative so we are left with $$x^2> \frac{1+ \sqrt{17}}{2}$$ which means $$x< -\sqrt{\frac{1+ \sqrt{17}}{2}}$$ and $$x> \sqrt{\frac{1+ \sqrt{17}}{2}}$$.

• In what way does this answer the question ? By the way, your resolution is wrong. – Yves Daoust Jun 15 '20 at 12:23
• I was lazy to write out where one function is smaller than other. – Milan Jun 15 '20 at 12:25