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How do you find the indefinite integral of #((sqrt(x) + (3/x) - 4 e^x)) dx#?

Answer 1

#2/3sqrt(x^3)+3lnx-4e^x+C#

Rewriting in a different form using laws of exponents and surds, and then integrating term by term using normal rules of integration, we get :

#int(x^(1/2)+3*1/x-4e^x)dx#

#=(x^(3/2))/(3/2)+3lnx-4e^x+C#

#=2/3sqrt(x^3)+3lnx-4e^x+C#

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Answer 2

Aurora Alvarado

To find the indefinite integral of ( \sqrt{x} + \frac{3}{x} - 4e^x ) with respect to ( x ), you integrate each term separately using the rules of integration:

For ( \sqrt{x} ), use the power rule for integration: ( \int x^n , dx = \frac{x^{n+1}}{n+1} + C ), where ( n \neq -1 ).
For ( \frac{3}{x} ), use the power rule for integration as well.
For ( e^x ), use the rule ( \int e^x , dx = e^x + C ).

Therefore, the indefinite integral of the given expression is:

[ \int \left( \sqrt{x} + \frac{3}{x} - 4e^x \right) , dx = \frac{2}{3}x^{\frac{3}{2}} + 3\ln|x| - 4e^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.