What is the integral of (x^4+x-36)/(x^2+6) ?

The integral of (x^4+x-36)/(x^2+6) is (x^3)/3-6x+1/2 ln|x^2+6|+C

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integral of (x^4+x-36)/(x^2+6)

\int \frac{x ^{4} + x - 36}{x ^{2} + 6} d x

\frac{x ^{3}}{3} - 6 x + \frac{1}{2} ln x^{2} + 6 + C

Solution steps

\int \frac{x ^{4} + x - 36}{x ^{2} + 6} d x

Long division

\frac{x ^{4} + x - 36}{x ^{2} + 6} : x^{2} - 6 + \frac{x}{x ^{2} + 6}

= \int x^{2} - 6 + \frac{x}{x ^{2} + 6} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{2} d x - \int 6 d x + \int \frac{x}{x ^{2} + 6} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 6 d x = 6 x

\int \frac{x}{x ^{2} + 6} d x = \frac{1}{2} ln x^{2} + 6

= \frac{x ^{3}}{3} - 6 x + \frac{1}{2} ln x^{2} + 6

Add a constant to the solution

= \frac{x ^{3}}{3} - 6 x + \frac{1}{2} ln x^{2} + 6 + C

11 x d y + 11 y d x + x d x = 0

\int_{2}^{4} \frac{1}{x ^{2}} d x

t^{2} y^{^{''}} + t y^{^{'}} + 16 y = 0

\int \frac{( 2 x - 1 ) ( x + 3 )}{6} d x

(0, 0), (2, 1)

What is the integral of (x^4+x-36)/(x^2+6) ?
The integral of (x^4+x-36)/(x^2+6) is (x^3)/3-6x+1/2 ln|x^2+6|+C