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

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

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integral of (-6x^4+6)/(x^2+1)

\int \frac{- 6 x ^{4} + 6}{x ^{2} + 1} d x

- 2 x^{3} + 6 x + C

Solution steps

\int \frac{- 6 x ^{4} + 6}{x ^{2} + 1} d x

Expand

\frac{- 6 x ^{4} + 6}{x ^{2} + 1} : - \frac{6 x ^{4}}{x ^{2} + 1} + \frac{6}{x ^{2} + 1}

Apply the Sum Rule:

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

= - \int \frac{6 x ^{4}}{x ^{2} + 1} d x + \int \frac{6}{x ^{2} + 1} d x

\int \frac{6 x ^{4}}{x ^{2} + 1} d x = 6 (\frac{x ^{3}}{3} - x + arctan (x))

\int \frac{6}{x ^{2} + 1} d x = 6 arctan (x)

= - 6 (\frac{x ^{3}}{3} - x + arctan (x)) + 6 arctan (x)

Simplify

- 6 (\frac{x ^{3}}{3} - x + arctan (x)) + 6 arctan (x) : - 2 x^{3} + 6 x

= - 2 x^{3} + 6 x

Add a constant to the solution

= - 2 x^{3} + 6 x + C

\int \frac{1}{e ^{10 x}} d x

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

\frac{d}{d x} (3 x^{- 3})

\int t^{2} - 2 t^{4} d t

f (x) = \frac{4 x + 3}{4 x - 3}

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