How do you evaluate the definite integral #4int x^2/(x-1)# from #[3,2]#?

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Answer 1 · 2017-03-07T00:18:17

#4int_3^2 x^2/(x-1)dx =-14-4ln2#

We can start by adding and subtracting #1# to the numerator:

#4int_3^2 x^2/(x-1)dx = -4 int_2^3 (x^2-1+1)/(x-1)dx = -4 int_2^3 ((x^2-1)/(x-1)+1/(x-1))dx #

Now factorize the numerator of the first addendum:

#x^2-1 =(x+1)(x-1)#

#4int_3^2 x^2/(x-1)dx = -4 int_2^3 (((x+1)(x-1))/(x-1)+1/(x-1))dx = -4 int_2^3 ((x+1)+1/(x-1))dx#

Using the linearity of the integral:

#4int_3^2 x^2/(x-1)dx = -4 int_2^3 xdx -4int_2^3 dx -4int_2^3 (1/(x-1))dx#

and these are all regular integrals:

#4int_3^2 x^2/(x-1)dx = [-2x^2 -4x -4ln abs(x-1)]_2^3#

#4int_3^2 x^2/(x-1)dx = -18 -12 -4ln2 + 8 +8 -4ln1 =-14-4ln2#