Integral of $\int_{1}^{x} \frac{1}{t^2+4}dt$.

Question

I am computing

$$\int_{1}^{x} \frac{1}{t^2+4}dt$$

I have firstly re-written the problem as: $$\frac{1}{4} \int_{1}^{x} \frac{1}{(\frac{t^2}{4}+1)}dt,$$ then by substitution of $u=\frac{t}{2},$ this simplifies to: $$\frac{1}{2}\int_{\frac{1}{2}}^{\frac{x}{2}} \frac{1}{(u^2+1)}$$ After this step, I was going to use substitution of say p=tan(u), and use trigonometry to simplify. However I have found the following online: $$\int \frac{u'(t)}{(1+(u(t))^2}dt=arctan(u).$$ I was wondering where this identity came from, so I could use/explain it in my answer, to save having to do the trigonometry.

Thanks :)

Answer 1

It is a consequence of that fact the$$(\forall x\in\mathbb{R}):\arctan'(x)=\frac1{1+x^2}$$and of the chain rule.

Answer 2

write $$4\left(\frac{t^2}{4}+1\right)=4\left(\left(\frac{t}{2}\right)^2+1\right)$$ and substitute $$u=\frac{t}{2}$$

Answer 3

Hint:

Let $t=2\tan u$, giving $dt=2(\tan^2u+1)\,du$.

Then

$$\int\frac{dt}{t^2+4}=\int\frac{2\tan^2u+2}{4\tan^2u+4}du=\frac u2.$$

Answer 4

Recall that if $g(x)$ is the inverse of $f(x)$, then $$\frac{dg}{dx} = \frac{1}{\frac{df}{dx}}$$

So since $y = \arctan x$ and $\tan y = x$, we have

\begin{align*} \frac{dy}{dx} &= \frac{1}{\frac{d}{dy}\tan y} \\ &= \frac 1{\sec^2 y} \\ &= \frac 1{1 + \tan^2 y} \\ &= \frac 1{1 + x^2} \end{align*}

So if $$\frac d{du} \arctan u = \frac {u'}{1 + u^2} \tag{Remember the chain rule!}$$ then $$\int \frac d{du} \arctan u = \int \frac {u'}{1 + u^2}$$ i.e. $$\arctan u = \int \frac {u'}{1 + u^2}$$