How do you evaluate the definite integral #int (3/x^2-1)dx# from [1,2]?

Question

Ezra Adams · Accepted Answer

I found: #1/2# Let us write our integral in a slight different way to make the integraton easier: #int3x^-2dx-int1dx=# now we integrate using our usual rules of integration to get: #=3x^(-2+1)/(-2+1)-x=-3x^-1-x=-3/x-x# we didn't add the constant because we now apply the limits of integration #[1,2]# evaluating our result for #x# in both points and subtracting the values: when #x=2# we get: #-3/2-2=-7/2# when #x=1# we get: #-3/1-1=-4# Subtract them: #(-7/2)-(-4)=(-7+8)/2=1/2# so we get that: #int_1^2(3/x^2-x)dx=1/2#

Emilia Aguilar · Answer

undefined To evaluate the definite integral ∫(3/(x^2 - 1))dx from 1 to 2, you can use the method of partial fraction decomposition to rewrite the integrand as a sum of simpler fractions.

The integrand can be expressed as 3/(x^2 - 1) = 3/((x - 1)(x + 1)).

Using partial fraction decomposition, we can write 3/((x - 1)(x + 1)) = A/(x - 1) + B/(x + 1), where A and B are constants.

By finding common denominators and equating coefficients, we find that A = 1 and B = -1.

So, the integral becomes ∫(1/(x - 1) - 1/(x + 1))dx from 1 to 2.

Integrating each term separately, we get ∫(1/(x - 1) - 1/(x + 1))dx = ln|x - 1| - ln|x + 1| evaluated from 1 to 2.

Substituting the limits of integration, we have ln|2 - 1| - ln|2 + 1| - (ln|1 - 1| - ln|1 + 1|).

Simplifying further, we get ln(1) - ln(3) - (ln(0) - ln(2)).

Since ln(0) is undefined, this term contributes -∞ to the integral.

Thus, the integral is evaluated as ln(1) - ln(3) - (-∞).

ln(1) is 0, and ln(3) is a finite number.

So, the integral evaluates to -∞ - 0 + ln(3).

Therefore, the definite integral ∫(3/(x^2 - 1))dx from 1 to 2 is -∞ + ln(3).