How do you find the definite integral of #int (x^2-2)/(x+1)# from #[0,2]#?

Question

Paisley Johnson · Accepted Answer

#-ln3#

Start by dividing the numerator by the denominator using long or synthetic division.

Thus:

#int_0^2(x^2 - 2)/(x + 1)dx = int_0^2 x - 1 + -1/(x + 1)dx#

We can integrate using the rules #int(1/x)dx= ln|x| + C# and #int(x^n)dx = x^(n +1)/(n + 1) + C#.

#= [1/2x^2 - x - ln|x + 1|]_0^2#

Evaluate using #int_a^b F(x) = f(b) - f(a)#, where #f'(x) = F(x)#.

#=1/2(2)^2 - 2 - ln|3| - (1/2(0)^2 - 0 - ln|0 + 1|)#

#=1/2(4) - 2 - ln|3| - 0#

#= -ln3#

In celebration of this being the 2000th answer I ever wrote for socratic, I've included a whole bunch of practice exercises for your improvement

Practice exercises

Evaluate each definite integral. Round answers to the nearest integer.
a) #int_1^4 (x^3 + 7x + 14)/(x + 2)dx#
b) #int_7^15 (2x^4 - 18x^3 + 2x^2 - 5x + 1)/(2x + 1)#
c) #int_3^6 (5x^5 + 2x^2 - 8)/(2(x + 4))#
Bonus
Hint: Use substitution
b) #int_e^(e + 3) (e^(2x) - e^x)/sqrt(e^x) dx#
Hopefully this helps!

Answers to practice exercises:
#1. 33#
#2. 2920#
#3. 2102#
bonus: #3474#

Adam Adkins · Answer

undefined To find the definite integral of $ \int_{0}^{2} \frac{x^2 - 2}{x + 1} \, dx $, you can use the method of partial fraction decomposition. After decomposing the fraction, integrate each term separately. Then, evaluate the result at the upper limit (2) and subtract the result evaluated at the lower limit (0).