How do you evaluate the integral #int (x^3-4)/(x+1)#?

Question

Ezekiel Alexander · Accepted Answer

#int(x^3-4)/(x+1)dx# Let #u=x+1#, implying that #du=dx#. This also means that #x=u-1#. Then: #=int((u-1)^3-4)/udu# Expand #(u-1)^3#: #=int((u^3-3u^2+3u-1)-4)/u# #=int(u^2-3u+3-5/u)du# Integrate term by term: #=1/3u^3-3/2u^2+3u-5lnabsu+C# #=1/3(x+1)^3-3/2(x+1)^2+3(x+1)-5lnabs(x+1)+C#

Scarlett Allison · Answer

undefined To evaluate the integral ∫(x^3 - 4)/(x + 1), perform polynomial long division to rewrite the integrand as x^2 - x + 1 - 3/(x + 1). Then, integrate each term separately. The integral of x^2 - x + 1 with respect to x is (1/3)x^3 - (1/2)x^2 + x, and the integral of -3/(x + 1) can be found using the natural logarithm. Therefore, the value of the integral is (1/3)x^3 - (1/2)x^2 + x - 3ln|x + 1| + C, where C is the constant of integration.

Ezekiel Alexander · Answer

undefined To evaluate the integral $ \int \frac{x^3 - 4}{x + 1} $, you can use polynomial long division or partial fraction decomposition. Here, I'll demonstrate how to use partial fraction decomposition:

1. First, factor the numerator $ x^3 - 4 $ if possible. Since it's a difference of cubes, you can factor it as $ (x - \sqrt[3]{4})(x^2 + \sqrt[3]{4}x + \sqrt[3]{16}) $.
2. Write the fraction as a sum of simpler fractions by decomposing it into partial fractions.
3. The degree of the numerator is 3, and the degree of the denominator is 1, so the decomposition will include both linear and constant terms.
4. Assume the decomposition is of the form $ \frac{x^3 - 4}{x + 1} = Ax^2 + Bx + C $.
5. Multiply both sides by $ x + 1 $ to clear the denominator.
6. Equate coefficients of corresponding powers of $ x $ on both sides.
7. Solve the resulting system of equations for $ A $, $ B $, and $ C $.
8. Once you find $ A $, $ B $, and $ C $, integrate each term separately.
9. The integral of the original function is the sum of the integrals of the decomposed terms.

By following these steps, you can evaluate the integral of $ \frac{x^3 - 4}{x + 1} $.