integral of 1/(x^2 - 9) dx ???

Answer 1

Integral (1/(x^2 - 9))dx

To solve this, we have to use trigonometric substitution. I'm going to use "t" instead of theta, though on paper I would normally use theta.

Let x = 3sec(t)
dx = 3sec(t)tan(t)dt

After the substitution, we have

Integral { ( 1 / [9sec^2(t) - 9] ) (3sec(t)tan(t)) } dt

Simplifying, we have

Integral ( [3sec(t)tan(t)] / [9sec^2(t) - 9] ) dt

Factoring out a 9 from the bottom, we have

Integral ( [3sec(t)tan(t)] / [9(sec^2(t) - 1)] )dt

By the identity "tan squared equals secant squared minus one",

Integral ( [3sec(t)tan(t)] / [9tan^2(t)] ) dt

Note that the 3 on the top will cancel out from the 9 in the bottom, leaving us with

Integral ( [sec(t)tan(t)] / [3tan^2(t)] ) dt

Let's pull out that 3 in the denominator. We would get a (1/3) outside of the integral as a result.

(1/3) * Integral ( [sec(t)tan(t)] / [tan^2(t)] ) dt

Note that the tan(t) on the numerator will cancel with the tan term in the denominator.

(1/3) * Integral ( [sec(t)]/[tan(t)] ) dt

Now, let's change everything to sines and cosines.

(1/3) * Integral ( [1/cos(t)] / [sin(t)/cos(t)] ) dt

Multiply the top and bottom by cos(t), we have

(1/3) * Integral ( 1/sin(t) ) dt

By definition,

(1/3) * Integral (csc(t)) dt

The integral of csc(t) is not something obvious, but definitely something one *needs* to know. The answer is
ln |csc(t) - cot(t)|.

(1/3) * ln |csc(t) - cot(t) | + C

At this point, we need to get the x variable in the solution.

Note that earlier, we assigned x = 3sec(t). From here, it follows that sec(t) = x/3

Remember SOHCAHTOA? Since cos(t) = adj/hyp, it follows that secant, which is the reciprocal of cosine, is hyp/adj.
Therefore, from sec(t) = x/3, construct a right angle triangle with angle t, and hypotenuse equal to x, adjacent equal to 3. It follows that the remaining side, the opposite, will be sqrt(x^2 - 9).

hyp = x
adj = 3
opp = sqrt(x^2 - 9)


(1/3) * ln |csc(t) - cot(t) | + C

csc(t) = hyp/opp = x/3
cot(t) = adj/opp = 3/sqrt(x^2 - 9), so our answer is

(1/3) * ln |x/3 - 3/sqrt(x^2 - 9)| + C

Answer 2

First graph the triangle and see: the base is 2 and height is 9 a) The formula for the area is (base)(height)/2 = 2*9/9 = 9 b)Other way is: find the equation of the line using the points (1,0) and(3,9)... it is y = (9/2)(x-1) Now use the integral: area = integral(from 1 to 3) of the function (9/2)(x-1) The integral of (9/2)(x-1) is (9/2)(x^2/2 - x) ... from 1 to 3.... the result is 9 OK!

Answer 3

You should discompose the integrand in simple fraction1/(x^2-9)= a/(x-3)+b/(x+3) For calculating a and b you can put:
1= a*(x+3 )+ b*(x-3) so a=1/6 and b=-1/6 so 1/(x^2-9) =
1/6* (1/(x+3) -1/(x-3) )
Integrating each summand you get 1/6 (logIx+3I - logIx-3I) which can be written as

logI(x+3)/(x-3)I^1/6 log means" e "based logs