Question

Consider the indefinite integral ∫x2−93x³+5x².−22x−42dx Then the integrand decomposes into the form ax+b+x−3c+x+3d where

a=
b=
c=
d=
Integrating term by term, we obtain that ∫x2−93x3+5x2−22x−42dx= +C

Answer 1

To decompose the given integrand, we equate the coefficients in the integrand with the coefficients in the decomposed form. Then, we integrate each term of the decomposed form separately, resulting in the indefinite integral (1/3)x^3 + (-93/4)x^4 + (5/3)x^3 - 11x^2 - 42x + C.

Step-by-step explanation:

To decompose the integrand, we need to manipulate it into the form ax+b+x-3c+x+3d. Comparing the given integrand x^2-9*3x^3+5x^2-22x-42dx to the decomposed form, we can equate the corresponding coefficients. By comparing the coefficients, we find that a=1, b=-93, c=5, and d=-22.

Next, we can integrate the decomposed form term by term. The indefinite integral of x^2 is (1/3)x^3, the indefinite integral of -93x^3 is (-93/4)x^4, the indefinite integral of 5x^2 is (5/3)x^3, the indefinite integral of -22x is (-11)x^2, and the indefinite integral of -42 is -42x. Adding all these integrals together, we get the indefinite integral of x^2-9*3x^3+5x^2-22x-42dx as (1/3)x^3 + (-93/4)x^4 + (5/3)x^3 - 11x^2 - 42x + C, where C is the constant of integration.