Evaluate the indefinite integral.∫(4ex+1x3)dx\int_{}^{}\left(4e^{x}+\frac{1}{x^3}\right)dx

4 e x − 1 2 x 2 + C 4e^{x}-\frac{1}{2x^2}+C

Evaluate the indefinite integral. ∫ ( 4 e x + 1 x 3 ) d x \int_{}^{}\left(4e^{x}+\frac{1}{x^3}\right)dx

we're asked to evaluate the indefinite integral of four E to the X plus one over X cubed DX. Now one over X cubed is the same thing as X to the power of negative three. So that's just going to help us later when we apply the general power rule. Now for this first term here, finding the antiderivative, four is just a constant. So using the constant multiple rule, this is really four times the antiderivative of e to the x, which is still e to the x. So I actually end up with the same exact function that I started with there, four e to the x. Then applying the general power rule to this next term, x to the power of negative three. This ends up giving us here a negative one half times x to the power of negative two. Then of course, adding on our constant of integration c. Now I can rewrite this slightly as four e to the x minus one over two x squared, just going ahead and putting that on the bottom because that exponent is negative, then adding on that c there and this is my final answer. Let us know if you have questions and let's keep practicing.

Evaluate the indefinite integral. ∫ ( 4 e x + 1 x 3 ) d x \int_{}^{}\left(4e^{x}+\frac{1}{x^3}\right)dx

4 e x − 1 2 x 2 + C 4e^{x}-\frac{1}{2x^2}+C

4 e x − 1 4 x 4 + C 4e^{x}-\frac{1}{4x^4}+C

4 x e x − 1 2 x 2 + C 4xe^{x}-\frac{1}{2x^2}+C

4 x e x − 1 4 x 4 + C 4xe^{x}-\frac{1}{4x^4}+C

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0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
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5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
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- Area Between Curves
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10. Physics Applications of Integrals 3h 16m
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11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
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14. Sequences & Series5h 36m
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- Introduction to Power Series
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16. Parametric Equations & Polar Coordinates7h 58m

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

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Multiple Choice

Evaluate the indefinite integral.
$\int_{}^{} (4 e^{x} + \frac{1}{x^{3}}) d x$

$4 e^{x} - \frac{1}{4 x^{4}} + C$

$4 x e^{x} - \frac{1}{2 x^{2}} + C$

$4 x e^{x} - \frac{1}{4 x^{4}} + C$

$4 e^{x} - \frac{1}{2 x^{2}} + C$

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Verified step by step guidance

Step 1: Break the integral into two separate terms for easier evaluation. The given integral is ∫(4e^x + 1/x^3) dx. This can be written as ∫4e^x dx + ∫(1/x^3) dx.

Step 2: Evaluate the first term, ∫4e^x dx. The integral of e^x is e^x, so multiplying by the constant 4 gives 4e^x.

Step 3: Evaluate the second term, ∫(1/x^3) dx. Rewrite 1/x^3 as x^(-3). Using the power rule for integration, ∫x^n dx = (x^(n+1))/(n+1) + C (for n ≠ -1), we get ∫x^(-3) dx = (x^(-3+1))/(-3+1) = x^(-2)/(-2) = -1/(2x^2).

Step 4: Combine the results of the two integrals. The result is 4e^x - 1/(2x^2).

Step 5: Add the constant of integration, C, to account for the indefinite integral. The final expression is 4e^x - 1/(2x^2) + C.