Evaluate the indefinite integral.∫(4ex+1x3)dx\int_{}^{}\left(4e^{x}+\frac{1}{x^3}\right)dx∫(4ex+x31)dx

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

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 + x 3 1 ) d x

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 + x 3 1 ) d x

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

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

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

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

10. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

Business Calculus

10. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

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Multiple Choice

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

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

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

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

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

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 rewritten 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 the integral of 4e^x is 4e^x. This gives us the first part of the solution.

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 integrate x^(-3). Adding 1 to the exponent gives x^(-2), and dividing by the new exponent (-2) gives -1/(2x^2).

Step 4: Combine the results from the two terms. The integral becomes 4e^x - 1/(2x^2) + C, where C is the constant of integration.

Step 5: Verify the solution by differentiating the result. Differentiate 4e^x - 1/(2x^2) + C to ensure it matches the original integrand, 4e^x + 1/x^3. This confirms the solution is correct.

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