Question

Find the area in the first quadrant under the curve y = 1 / (x^2+6x+10)

Answer 1

The area is given by

$\displaystyle\int_0^\infty\frac{\mathrm dx}{x^2+6x+10}$

Complete the square in the denominator:

$x^2+6x+10=x^2+6x+9+1=(x+3)^2+1$

then substitute $x+3=\tan y$, so that $\mathrm dx=\sec^2y\,\mathrm dy$. Then as $x\to0^+$, we have $y\to\arctan3$; as $x\to\infty$, we have $y\to\dfrac\pi2$. So the integral becomes

$\displaystyle\int_{\arctan3}^{\pi/2}\frac{\sec^2y}{\tan^2y+1}\,\mathrm dy=\int_{\arctan3}^{\pi/2}\mathrm dy=\dfrac\pi2-\arctan3$