How do you find the integral of #int (x^4-1)/(x^2+1)dx#?
We have that
#int (x^4-1)/(x^2+1)dx=int [(x^2+1)*(x^2-1)]/(x^2+1)dx= int (x^2-1)dx=x^3/3-x+c#
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To find the integral of (\int \frac{x^4 - 1}{x^2 + 1} , dx), you can use polynomial long division to rewrite the integrand as (x^2 - 1) plus a proper fraction. Then, integrate each term separately.
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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.
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.
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.
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.
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