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How do you integrate #(x^2+1)(x-1) dx#?

Answer 1

Ezra Adams

Expand the multiplication, then integrate term by term.

#int(x^2+1)(x-1) dx = int (x^3-x^2+x-1) dx#

# = 1/4x^4-1/3x^3+1/2x^2-1 +C#

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Answer 2

William Abdalla

To integrate ( (x^2 + 1)(x - 1) ) with respect to ( x ), you can expand the expression and then integrate each term separately:

( (x^2 + 1)(x - 1) = x^3 - x^2 + x - 1 )

Now, integrate each term:

( \int x^3 dx - \int x^2 dx + \int x dx - \int 1 dx )

( = \frac{1}{4}x^4 - \frac{1}{3}x^3 + \frac{1}{2}x^2 - x + C )

Where ( C ) is the constant of integration.

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Answer 3

Axel Alvarez

To integrate ((x^2 + 1)(x - 1) , dx), you can expand the expression and then integrate each term individually.

((x^2 + 1)(x - 1) = x^3 - x^2 + x - 1)

Now, integrate each term:

(\int (x^3 - x^2 + x - 1) , dx = \frac{1}{4}x^4 - \frac{1}{3}x^3 + \frac{1}{2}x^2 - x + C)

So, the integral of ((x^2 + 1)(x - 1) , dx) is (\frac{1}{4}x^4 - \frac{1}{3}x^3 + \frac{1}{2}x^2 - x + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.