Which line is perpendicular to the line whose equation is 5y + 6 = -3x?

Mathematics

1 answer:

Lynna [10]3 years ago

6 0

Answer: B

Step by Step

5y + 6 = -3X

5y = -3X - 6

Y = -3/5X -6/5

Perpedicular is the flip of your slope so -3/5 becomes 5/3X

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Integral of x"2+4/x"2+4x+3

dolphi86 [110]

I'm guessing you mean

$\displaystyle \int\frac{x^2+4}{x^2+4x+3}\,\mathrm dx$

First, compute the quotient:

$\displaystyle \frac{x^2+4}{x^2+4x+3} = 1 + \frac{4x-1}{x^2+4x+3}$

Split up the remainder term into partial fractions. Notice that

x ² + 4x + 3 = (x + 3) (x + 1)

Then

$\displaystyle \frac{4x-1}{x^2+4x+3} = \frac a{x+3} + \frac b{x+1} \\\\ \implies 4x - 1 = a(x+1) + b(x+3) = (a+b)x + a+3b \\\\ \implies a+b=4 \text{ and }a+3b = -1 \\\\ \implies a=\frac{13}2\text{ and }b=-\frac52$

So the integral becomes

$\displaystyle \int \left(1 + \frac{13}{2(x+3)} - \frac{5}{2(x+1)}\right) \,\mathrm dx = \boxed{x + \frac{13}2\ln|x+3| - \frac52 \ln|x+1| + C}$

We can simplify the result somewhat:

$\displaystyle x + \frac{13}2\ln|x+3| - \frac52 \ln|x+1| + C \\\\ = x + \frac12 \left(13\ln|x+3| - 5\ln|x+1|\right) + C \\\\ = x + \frac12 \left(\ln\left|(x+3)^{13}\right| - \ln\left|(x+1)^5\right|\right) + C \\\\ = x + \frac12 \ln\left|\frac{(x+3)^{13}}{(x+1)^5}\right| + C \\\\ = \boxed{x + \ln\sqrt{\left|\frac{(x+3)^{13}}{(x+1)^5}\right|} + C}$