What is the integral of (x^6+2x^4+6x-9)/(x^3+3) ?

The integral of (x^6+2x^4+6x-9)/(x^3+3) is 1/4 x^4+x^2-3x-1/3+C

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integral of (x^6+2x^4+6x-9)/(x^3+3)

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

\frac{1}{4} x^{4} + x^{2} - 3 x - \frac{1}{3} + C

Solution steps

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

Apply Integration By Parts

: x (x - 1) (x^{2} + x + 3) - \int x (3 x^{2} + 2) d x

= x (x - 1) (x^{2} + x + 3) - \int x (3 x^{2} + 2) d x

\int x (3 x^{2} + 2) d x = \frac{1}{12} (3 x^{2} + 2)^{2}

= x (x - 1) (x^{2} + x + 3) - \frac{1}{12} (3 x^{2} + 2)^{2}

Simplify

x (x - 1) (x^{2} + x + 3) - \frac{1}{12} (3 x^{2} + 2)^{2} : \frac{1}{4} x^{4} + x^{2} - 3 x - \frac{1}{3}

= \frac{1}{4} x^{4} + x^{2} - 3 x - \frac{1}{3}

Add a constant to the solution

= \frac{1}{4} x^{4} + x^{2} - 3 x - \frac{1}{3} + C

\frac{6}{( 1 - x ) ^{3}}

k \to \infty lim ((- 1)^{k})

x \to \infty lim (7 x^{2})

f^{^{'}} (x) = (3 x^{6} + 4 x^{3})^{4}

x^{3} + 8, at x = - 2

What is the integral of (x^6+2x^4+6x-9)/(x^3+3) ?
The integral of (x^6+2x^4+6x-9)/(x^3+3) is 1/4 x^4+x^2-3x-1/3+C