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djyliett [7]
2 years ago
5

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

Mathematics
1 answer:
dolphi86 [110]2 years ago
3 0

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

<em>x</em> ² + 4<em>x</em> + 3 = (<em>x</em> + 3) (<em>x</em> + 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}

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#1) A. 4; #2) A. periodic; about 6; #3) B. not periodic; #4) C.  2; 0.5; #5) A. 0.05 seconds; 4.5.

Explanation:

#1)  The period of a function is essentially the amount of time it takes for the function to start all over and repeat itself.  In this function, at t = 0 the graph is at 1; it curves up, back down and begins again at y=1 when t=4.  This means the period goes from t=0 to t=4, so it is 4.

#2) Looking at the left side of the graph, specifically the peak at (-5, 2), we see the same peak at about (1, 2).  Following the graph after that we can see that it does indeed repeat itself; this means the period goes from t= - 5 to t = 1, so it is 6.

#3) This function never repeats, so it is not periodic.

#4) This function repeats when it reaches t=2, so 2 is the period.  The amplitude is the distance from the center line (of the graph, not the x-axis) to the peak.  The center line would be located at about y=0.5; the peaks are at y=-1.  This means the amplitude is 0.5.

#5) This function repeats every 0.05 seconds.  In this case, the center line is the x-axis; the distance from it to any peak is 4.5, so 4.5 is the amplitude.

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