Question

Antiderivative of a square root1" title=" \sqrt{1 + t ^{2} }" alt=" \sqrt{1 + t ^{2} }" align="absmiddle" class="latex-formula">
​

Answer 1

Substitute $t=\tan s$, so that $\mathrm dt=\sec^2s\,\mathrm ds$. The integral is then equivalent to

$\displaystyle\int\sqrt{1+(\tan s)^2}\,\sec^2s\,\mathrm ds$

$=\displaystyle\int\sqrt{1+\tan^2s}\,\sec^2s\,\mathrm ds$

$=\displaystyle\int\sqrt{\sec^2s}\,\sec^2s\,\mathrm ds$

In general, $\sqrt{x^2}=|x|$, so $\sqrt{\sec^2s}=|\sec s|$.

We want the substitution made above to be reversible, so that $s=\tan^{-1}t$. This restricts $s$ to the interval $-\dfrac\pi2$, and over this interval we have $\cos s>0\implies\sec s>0$, so we take the positive square root in order that $\sqrt{\sec^2s}=+\sec s$.

Then the integral becomes

$=\displaystyle\int\sec^3s\,\mathrm ds$

which can be computed in several ways. One method is to integrate by parts, taking

$u=\sec s\implies\mathrm du=\sec s\tan s\,\mathrm ds$

$\mathrm dv=\sec^2s\,\mathrm ds\implies v=\tan s$

so that

$\displaystyle\int\sec^3s\,\mathrm ds=\sec s\tan s-\int\sec s\tan^2s\,\mathrm ds$

$\displaystyle\int\sec^3s\,\mathrm ds=\sec s\tan s-\int\sec s(\sec^2s-1)\,\mathrm ds$

$\displaystyle\int\sec^3s\,\mathrm ds=\sec s\tan s-\int(\sec^3s-\sec s)\,\mathrm ds$

$\displaystyle2\int\sec^3s\,\mathrm ds=\sec s\tan s+\int\sec s\,\mathrm ds$

$\displaystyle\int\sec^3s\,\mathrm ds=\frac12\sec s\tan s+\frac12\ln|\sec s+\tan s|+C$

Then with $s=\tan^{-1}t$, you have $\tan s=t$ and $\sec s=\sqrt{1+t^2}$, which follows from the Pythagorean identity. So

$\displaystyle\int\sqrt{1+t^2}\,\mathrm dt=\frac{t\sqrt{1+t^2}+\ln|t+\sqrt{1+t^2}|}2+C$