Evaluate int(x^(6)+2x^(4)+6x-9)/(x^(3)+3)dx Choose 1 answer: (A) (x^(4))/(4)+x^(2)-3x+C (B) (x^(4))/(4)+x^(2)-9x+C (C) (x^(4))/(4)+3x^(2)+3x+C (D) (x^(4))/(4)+x^(2)+3x+C

Divide and Sum Integrals: Divide x^6 by x^3 to get x^3, and 2x^4 by x^3 to get 2x. The integral becomes the sum of the integrals of x^3, 2x, 6x/(x^3+3), and -9/(x^3+3).Integrate x^3: Integrate x^3 to get (x^4)/4.Integrate 2x: Integrate 2x to get x^2.Integrate 6x/(x^3+3): Integrate 6x/(x^3+3). This is a bit tricky, but notice that the derivative of x^3+3 is 3x^2, which is close to 6x. Let's use substitution: let u = x^3+3, then du = 3x^2 dx. We need to adjust for the 6x, so we multiply du by 2/3 to get the 6x. The integral of 6x/(x^3+3) becomes 2 times the integral of du/u.Integrate 2*du/u: Integrate 2*du/u to get 2ln|u|.Substitute u back: Substitute back for u to get 2ln|x^3+3|.Integrate -9/(x^3+3): Integrate -9/(x^3+3). This is a constant over a linear term, so it's just -9 times the integral of 1/(x^3+3). This doesn't have an elementary antiderivative, so it looks like I made a mistake here. The integral of -9/(x^3+3) should be treated as a separate term.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Evaluate
int(x^(6)+2x^(4)+6x-9)/(x^(3)+3)dx
Choose 1 answer:
(A)
(x^(4))/(4)+x^(2)-3x+C
(B)
(x^(4))/(4)+x^(2)-9x+C
(C)
(x^(4))/(4)+3x^(2)+3x+C
(D)
(x^(4))/(4)+x^(2)+3x+C

Evaluate $\int \frac{x^{6}+2 x^{4}+6 x-9}{x^{3}+3} d x$ $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{x^{4}}{4}+x^{2}-3 x+C$ $\newline$ (B) $\frac{x^{4}}{4}+x^{2}-9 x+C$ $\newline$ (C) $\frac{x^{4}}{4}+3 x^{2}+3 x+C$ $\newline$ (D) $\frac{x^{4}}{4}+x^{2}+3 x+C$

View step-by-step help

Home

Math Problems

Calculus

Evaluate definite integrals using the chain rule

Full solution

Q. Evaluate

\int \frac{x^{6}+2 x^{4}+6 x-9}{x^{3}+3} d x

\newline

Choose

1

answer:

\newline

(A)

\frac{x^{4}}{4}+x^{2}-3 x+C

\newline

(B)

\frac{x^{4}}{4}+x^{2}-9 x+C

\newline

(C)

\frac{x^{4}}{4}+3 x^{2}+3 x+C

\newline

(D)

\frac{x^{4}}{4}+x^{2}+3 x+C

Divide and Sum Integrals: Divide $x^6$ by $x^3$ to get $x^3$ , and $2x^4$ by $x^3$ to get $2x$ . The integral becomes the sum of the integrals of $x^3$ , $2x$ , rac{6x}{x^3+3}, and -rac{9}{x^3+3}.
Integrate $x^3$ : Integrate $x^3$ to get $\frac{x^4}{4}$ .
Integrate $2x$ : Integrate $2x$ to get $x^2$ .
Integrate $\frac{6x}{x^3+3}$ : Integrate $\frac{6x}{x^3+3}$ . This is a bit tricky, but notice that the derivative of $x^3+3$ is $3x^2$ , which is close to $6x$ . Let's use substitution: let $u = x^3+3$ , then $du = 3x^2 dx$ . We need to adjust for the $6x$ , so we multiply $du$ by $\frac{2}{3}$ to get the $6x$ . The integral of $\frac{6x}{x^3+3}$ becomes $\frac{6x}{x^3+3}$ $2$ times the integral of $\frac{6x}{x^3+3}$ $3$ .
Integrate $2\frac{du}{u}$ : Integrate $2\frac{du}{u}$ to get $2\ln|u|.$
Substitute $u$ back: Substitute back for $u$ to get $2\ln|x^3+3|$ .
Integrate $-9/(x^3+3)$ : Integrate $-9/(x^3+3)$ . This is a constant over a linear term, so it's just $-9$ times the integral of $1/(x^3+3)$ . This doesn't have an elementary antiderivative, so it looks like I made a mistake here. The integral of $-9/(x^3+3)$ should be treated as a separate term.