Interesting Integrals 9

Relevant wiki: Trigonometric Substitution in Integration

Substitute $x=\tan y$ so that $\mathrm{d}x=\sec^2y ~\mathrm{d}y=(1+\tan^2 y)~\mathrm{d}y$ and integration is:

$\large \mathfrak G=\displaystyle\int_{\pi/4}^{\pi/3}\dfrac{ (1+\tan^2 y)~\mathrm{d}y}{ \sqrt{(1+\tan^2 y)^3}}$

$\large = \displaystyle\int_{\pi/4}^{\pi/3}\dfrac{ ~\mathrm{d}y}{\underbrace{ \sqrt{1+\tan^2 y}}_{\color{#D61F06}{\sqrt{\sec^2 y}=\sec y}}}$

$\large = \displaystyle\int_{\pi/4}^{\pi/3} \cos y~\mathrm{d}y$

$\large =\left[\sin y\right]_{\pi/4}^{\pi/3}=\dfrac{\sqrt 3-\sqrt 2}2$

Hence, $\Large 2+3=\boxed{\color{#0C6AC7}{5}}$

\begin{aligned} I & = \int_1^\sqrt 3 \frac {dx}{\sqrt{(1+x^2)^3}} & \small \color{#3D99F6}{\text{Let } x = \tan \theta \implies dx = \sec^2 \theta \ d \theta} \\ & = \int_\frac \pi 4 ^\frac \pi 3 \frac {\sec^2 \theta}{\sqrt{(\sec^2 \theta)^3}} d \theta \\ & = \int_\frac \pi 4 ^\frac \pi 3 \frac {\sec^2 \theta}{\sec^3 \theta} d \theta \\ & = \int_\frac \pi 4 ^\frac \pi 3 \frac 1{\sec \theta} d \theta \\ & = \int_\frac \pi 4 ^\frac \pi 3 \cos \theta \ d \theta \\ & = \sin \theta \ \bigg|_\frac \pi 4 ^\frac \pi 3 \\ & = \frac {\sqrt 3}2 - \frac {\sqrt 2}2 \\ & = \frac {\sqrt 3 - \sqrt 2}2 \end{aligned}

$\implies a+b = 3+2 = \boxed{5}$