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[tex]\displaystyle \int_1^3 \frac{\mathrm dx}{x^2-2x+5}[/tex]
Follow the instruction and complete the square in the denominator:
x ² - 2x + 5 = (x ² - 2x + 1) + 4 = (x - 1)² + 4
Then the integral is
[tex]\displaystyle \int_{x=1}^{x=3} \frac{\mathrm dx}{(x-1)^2+4}[/tex]
Substitute y = x - 1 and dy = dx :
[tex]\displaystyle \int_{y+1=1}^{y+1=3} \frac{\mathrm dy}{y^2+4} = \int_{y=0}^{y=2}\frac{\mathrm dy}{y^2+4}[/tex]
Substitute y = 2 tan(z) and dy = 2 sec²(z) dz :
[tex]\displaystyle \int_{2\tan(z)=0}^{2\tan(z)=2}\frac{2\sec^2(z)}{(2\tan(z))^2+4}\,\mathrm dz = \frac12 \int_{z=0}^{z=\pi/4} \frac{\sec^2(z)}{\tan^2(z)+1}\,\mathrm dz \\\\ = \frac12 \int_{z=0}^{z=\pi/4} \frac{\sec^2(z)}{\sec^2(z)}\,\mathrm dz \\\\ = \frac12 \int_{z=0}^{z=\pi/4} \mathrm dz = \frac12 z\bigg|_{z=0}^{z=\pi/4} = \frac12 \left(\frac\pi4-0\right) = \boxed{\frac\pi8}[/tex]