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How do you integrate #((x^3 + 7) / x^2) dx#?

Answer 1

Charles Anctil

#int((x^3+7)/x^2)dx= frac (x^3-14) (2x)#

#int((x^3+7)/x^2)dx = int (x^3/x^2+7/x^2)dx = int xdx+7int x^(-2)dx #

#= x^2/2-7x^-1 = x^2/2-7/x = frac (x^3-14) (2x)#

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Answer 2

Elijah Abram

To integrate (\frac{{x^3 + 7}}{{x^2}}) with respect to (x), you can use polynomial long division or rewrite the expression as (x + \frac{7}{{x^2}}) and integrate each term separately. Here's the step-by-step process:

Divide (x^3) by (x^2) to get (x).
Integrate (x) with respect to (x) to get (\frac{{x^2}}{2}).
For the term (\frac{7}{{x^2}}), rewrite it as (7x^{-2}).
Integrate (7x^{-2}) with respect to (x) to get (7 \cdot \frac{{x^{-1}}}{(-1)} = -\frac{7}{x}).

Combining the results from steps 2 and 4, the integral of (\frac{{x^3 + 7}}{{x^2}}) with respect to (x) is:

[\int \frac{{x^3 + 7}}{{x^2}} , dx = \frac{{x^2}}{2} - \frac{7}{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.