Calc1231W5T2 - Integrals of rational functions

Friday, 24 August 2018

1:05 PM

Untitled picture.png Machine generated alternative text:
The integral of a function of the form 
Hence 
5 
dc = In(aös(5x+2)) 
Note: the Maple syntax for In(lxl) is In(abs (x) ) 
1 
dc 
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Untitled picture.png Machine generated alternative text:
The integral 
1 
1 
dc. 
dc , as in the previous questions. But with a bit of algebraic manipulation we can make it into two integrals of this form. This is because 
is not in the form 
f@) 
1 
for some constants A and B 
Let's find the value of these constants. First multiply both sides of this equation by @ — 4)(x — 3) . This gives 
• Setting x 
• Setting x 
4 in this equation means that A 
3 in this equation means that B 
Now the functions 
are in the required form 
x—4' x—3 
This makes the original integral I easy to calculate as the sum of 
f@) 
and 
1 
dc 
1 
dc 
In(abs(x-4)) 


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Untitled picture.png Machine generated alternative text:
What about an integral of the form 
1 
1 
The plan is the same, but the execution is slightly different. We split up the integrand into three functions that we can integrate individually: 
1 
@ — l)(x — 
where A, B and C are constants. Ifwe multiply this equation by @ — 1)@ — then 
— + B(x — 1)@ — 4) + ¯ 1) 
Now lets find the values of these constants: 
• Setting x = 1 gives A = 1/9 
• Setting x = 4 gives C = 1/3 
We have to work a little harder to find B. Comparing coefficients of 
Hence our original integral I is the sum of the three integrals: 
1 
1 
dc 
1 
dc 
is probably the easiest way of finding B, since we only have to solve 
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Untitled picture.png Machine generated alternative text:
Using the techniques from this topic, it is possible to evaluate integrals such as 
dc = 
J +25 
and 
dc 
J x2 — 25 
0åt+C 
Untitled picture.png Machine generated alternative text:
It's not always possible to factorize a quadratic into real linear terms. For example 
does not factorize into real linear terms since the discriminant 
is less than zero. In this case we complete the square to write p@) as a sum of two squares: 
where 
Hence we can integrate 
by first completing the square 
Then using the standard integral formula 
we calculate 
Note: the discriminant of ax2 + bx c is A = b2 
4 
+ -4m + 20 
1 
dc 
1 
1 
1 
.T2 + 4X + 20 
1 
du 
2 
4ac 
1 
1 
and the Maple syntax for tan¯ 
1 
I(au) 
is arctan (x) 
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