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After I get form of

$\int \frac{x^2}{x+1}dx-\int \frac{2x}{x+1}dx-\int \frac{3}{x+1}dx$

then get

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

(problem here i think ^)

using limits; I get b-a

I'm getting $-2$

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

is something wrong with the equation above?

  • (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. –  Oct 21 '16 at 15:35

2 Answers2

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Hint:

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

Bernard
  • 175,478
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$$\int _{ 2 }^{ 3 } \: \frac { x^{ 2 }-2x-3 }{ x+1 } dx=\int _{ 2 }^{ 3 } \: \frac { \left( x-3 \right) \left( x+1 \right) }{ x+1 } dx=\int _{ 2 }^{ 3 } \: \left( x-3 \right) dx={ \left( \frac { { x }^{ 2 } }{ 2 } -3x \right) }_{ 2 }^{ 3 }=-\frac { 1 }{ 2 } $$

haqnatural
  • 21,578
  • ok you can factor these $\frac{\left(x-3\right)\left(x+1\right)}{x+1}$ and let say not these ? for example $\frac{\left(x-3\right)\left(x-1\right)}{x-1}$ –  Oct 21 '16 at 15:41
  • im confused what types you can factor when doing definite integral –  Oct 21 '16 at 15:42
  • to tell the truth i didn't understant your comment,can you clearify your point? – haqnatural Oct 21 '16 at 15:45
  • 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. –  Oct 21 '16 at 15:50
  • @TemurBakhriddinov,it is ok.This is another question.In your case, we can factor numerator – haqnatural Oct 21 '16 at 15:55
  • You can simplify in this case because $(x+1) = 0 \iff x=-1$, but that does not worry us here because we are integrating from $x=2$ to $x=3$, and on that interval, $(x+1)$ is never equal to 0. – amWhy Oct 21 '16 at 16:07
  • @amWhy while that's true, it doesn't really address the OP's concern. It should be noted that it wouldn't matter if the integration went across that point.... The Riemann Integral (which I assume the OP is using) can handle a finite number of discontinuities – Brevan Ellefsen Oct 22 '16 at 02:14