How do you integrate:
1/(x^2-1) dx
?
draw a right triangle, x as the hypotensue, 1 as the adjacent side to angle theta, and the opposite side is sqrt (x^2-1)
costheta=1/x
-sintheta* dT=-1/x^2 dx or
dx= x^2 sinTheta dT
the integral is now...
INt 1/tanTheta*x^2 sinTheta dT
INT cosT/SinT* sec^2T*sinT dT
INT= secT dT=ln(secT+tanT)
= ln(x+x^2-1)
check all that, I did it in my head.
Hm, I've never done integrals using that method before. I'll check it.
To integrate the function 1/(x^2-1) dx, we can use the method of partial fractions. This involves decomposing the rational function into simpler fractions that are easier to integrate.
First, let's factor the denominator, x^2 - 1. It can be written as (x - 1)(x + 1).
Next, we can write the integrand as a sum of two fractions with unknown coefficients A and B:
1/(x^2 - 1) = A/(x - 1) + B/(x + 1).
To find the values of A and B, we need to find a common denominator on the right-hand side:
1/(x^2 - 1) = (A(x + 1) + B(x - 1))/((x - 1)(x + 1)).
Now, we can equate the numerators:
1 = (A(x + 1) + B(x - 1)).
Expanding and collecting like terms:
1 = (A + B)x + (A - B).
Since this equation holds for all x, we can equate the coefficients of like powers of x:
A + B = 0 ---> A = -B,
and
A - B = 1.
Solving the above system of equations, we find A = 1/2 and B = -1/2.
Therefore, the partial fraction decomposition becomes:
1/(x^2 - 1) = 1/2(x - 1) - 1/2(x + 1).
Now, we can integrate each term separately:
∫(1/(x^2 - 1)) dx = ∫(1/2(x - 1) - 1/2(x + 1)) dx.
Integrating each term:
= (1/2)∫(1/(x - 1)) dx - (1/2)∫(1/(x + 1)) dx.
The integral of 1/(x - 1) is ln| x - 1|, and the integral of 1/(x + 1) is ln| x + 1|. Therefore, the final integral is:
= (1/2)ln| x - 1| - (1/2)ln| x + 1| + C,
where C is the constant of integration.
Therefore, the result of integrating 1/(x^2 - 1) dx is (1/2)ln| x - 1| - (1/2)ln| x + 1| + C.