# ∫32×2−2x−3x+1dx\int _2^3\:\frac{x^2-2x-3}{x+1}dx

After I get form of

∫x2x+1dx−∫2xx+1dx−∫3x+1dx\int \frac{x^2}{x+1}dx-\int \frac{2x}{x+1}dx-\int \frac{3}{x+1}dx

then get

(x+1)2−2(x+1)+ln|x+1|−2(x+1−ln|x+1|)−3ln|x+1|\left(x+1\right)^2-2\left(x+1\right)+ln\left|x+1\right|-2\left(x+1-ln\left|x+1\right|\right)-3ln\left|x+1\right|

(problem here i think ^)

using limits; I get b-a

I’m getting −2-2

though the answer is supposed to be −12-\frac{1}{2}

is something wrong with the equation above?

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(1) How did you get that expression for the antiderivative? (2) If you want to evaluate the integral the easy way, factor the numerator and simplify the rational expression.
– Bye_World
2 days ago

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2

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∫32×2−2x−3x+1dx=∫32(x−3)(x+1)x+1dx=∫32(x−3)dx=(x22−3x)32=−12\int _{ 2 }^{ 3 } \: \frac { x^{ 2 }-2x-3 }{ x+1 } dx=\int _{ 2 }^{ 3 } \: \frac { \left( x-3 \right) \left( x+1 \right) }{ x+1 } dx=\int _{ 2 }^{ 3 } \: \left( x-3 \right) dx={ \left( \frac { { x }^{ 2 } }{ 2 } -3x \right) }_{ 2 }^{ 3 }=-\frac { 1 }{ 2 }

ok you can factor these (x−3)(x+1)x+1\frac{\left(x-3\right)\left(x+1\right)}{x+1} and let say not these ? for example (x−3)(x−1)x−1\frac{\left(x-3\right)\left(x-1\right)}{x-1}
– Temur Bakhriddinov
2 days ago

im confused what types you can factor when doing definite integral
– Temur Bakhriddinov
2 days ago

to tell the truth i didn’t understant your comment,can you clearify your point?
– haqnatural
2 days ago

sorry about that ; pretty much am I able to factor and simplify any factorable ration-able expression? because I remember my professor saying that there are some that you can’t factor where you could have factored only in algebraic terms such as here.
– Temur Bakhriddinov
2 days ago

@TemurBakhriddinov,it is ok.This is another question.In your case, we can factor numerator
– haqnatural
2 days ago

Hint:

x2−2x−3=(x+1)(x−3)x^2-2x-3=(x+1)(x-3), so you can simplify the integrand.

@haqnatural: Oh! Yes.Thanks for pointing it. Fixed.
– Bernard
2 days ago