Can someone explain this to me since I don’t get it that well?

Mathematics

1 answer:

Tatiana [17]3 years ago

4 0

so it'd be the first, third and fourth one. all you gotta do is draw a vertical line through the middle of the 1st, 3rd and 4th.

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Read 2 more answers

∫<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%2B2019%7D%7Bx%5E%7B2%7D%2B9%20%7D" id="TexFormula1" title="\frac{x+2019}{x^{2}+9

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Split up the integral:

$\displaystyle\int\frac{x+2019}{x^2+9}\,\mathrm dx = \int\frac{x}{x^2+9}\,\mathrm dx + \int\frac{2019}{x^2+9}\,\mathrm dx$

For the first integral, substitute y = x ² + 9 and dy = 2x dx. For the second integral, take x = 3 tan(z) and dx = 3 sec²(z) dz. Then you get

$\displaystyle \int\frac x{x^2+9}\,\mathrm dx = \frac12\int{2x}{x^2+9}\,\mathrm dx \\\\ = \frac12\int\frac{\mathrm du}u \\\\ = \frac12\ln|u| + C \\\\ =\frac12\ln\left(x^2+9\right)$

and

$\displaystyle \int\frac{2019}{x^2+9}\,\mathrm dx = 2019\int\frac{3\sec^2(z)}{(3\tan(z))^2+9}\,\mathrm dz \\\\ = 2019\int\frac{3\sec^2(z)}{9\tan^2(z)+9}\,\mathrm dz \\\\ = 673\int\frac{\sec^2(z)}{\tan^2(z)+1}\,\mathrm dz \\\\ = 673\int\frac{\sec^2(z)}{\sec^2(z)}\,\mathrm dz \\\\ = 673\int\mathrm dz \\\\ = 673z+C \\\\ = 673\arctan\left(\frac x3\right)+C$

Then

$\displaystyle\int\frac{x+2019}{x^2+9}\,\mathrm dx = \boxed{\frac12\ln\left(x^2+9\right) + 673\arctan\left(\frac x3\right) + C}$

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