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How do you integrate #(x^3+2x^2-x) / x dx#?

Answer 1

Aurora Alvarado

#x^3/3+x^2-x+C#

At first we simplify the integrand

#(x^3+2x^2-x)/x=x^3/x+2x^2/x-x/x=x^2+2x-1# Now we can integrate the result:

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

we have used that

#int x^ndx=x^(n+1)/(n+1)+C# if #n ne -1#

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Answer 2

Ezekiel Alexander

To integrate ((x^3 + 2x^2 - x) / x) with respect to (x), you can first simplify the expression by dividing every term by (x), resulting in (x^2 + 2x - 1). Then, integrate each term separately. The integral of (x^2) is ((1/3)x^3), the integral of (2x) is (x^2), and the integral of (-1) is (-x). So, the integral of the given expression is ((1/3)x^3 + 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.