Question

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

Answer 1

`-ln3`

Explanation:

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

1. 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`

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