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

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

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integral of (x^4+x+4)/(x^2+2)

\int \frac{x ^{4} + x + 4}{x ^{2} + 2} d x

\frac{x ^{3}}{3} - 2 x + \frac{1}{2} ln x^{2} + 2 + 42 arctan (\frac{x}{2}) + C

Solution steps

\int \frac{x ^{4} + x + 4}{x ^{2} + 2} d x

Expand

\frac{x ^{4} + x + 4}{x ^{2} + 2} : \frac{x ^{4}}{x ^{2} + 2} + \frac{x}{x ^{2} + 2} + \frac{4}{x ^{2} + 2}

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{x ^{4}}{x ^{2} + 2} d x + \int \frac{x}{x ^{2} + 2} d x + \int \frac{4}{x ^{2} + 2} d x

\int \frac{x ^{4}}{x ^{2} + 2} d x = \frac{x ^{3}}{3} - 2 x + 22 arctan (\frac{x}{2})

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

\int \frac{4}{x ^{2} + 2} d x = 22 arctan (\frac{x}{2})

= \frac{x ^{3}}{3} - 2 x + 22 arctan (\frac{x}{2}) + \frac{1}{2} ln x^{2} + 2 + 22 arctan (\frac{x}{2})

Simplify

= \frac{x ^{3}}{3} - 2 x + \frac{1}{2} ln x^{2} + 2 + 42 arctan (\frac{x}{2})

Add a constant to the solution

= \frac{x ^{3}}{3} - 2 x + \frac{1}{2} ln x^{2} + 2 + 42 arctan (\frac{x}{2}) + C

4^{x}

\int \frac{11 x ^{2} + 5 x - 4}{x ^{3} + x ^{2} - 2 x} d x

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