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

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

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integral of (x^3+x+1)/(x^2+4)

\int \frac{x ^{3} + x + 1}{x ^{2} + 4} d x

\frac{1}{2} x^{2} + 2 - \frac{3}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{x ^{3} + x + 1}{x ^{2} + 4} d x

Expand

\frac{x ^{3} + x + 1}{x ^{2} + 4} : \frac{x ^{3}}{x ^{2} + 4} + \frac{x}{x ^{2} + 4} + \frac{1}{x ^{2} + 4}

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 ^{3}}{x ^{2} + 4} d x + \int \frac{x}{x ^{2} + 4} d x + \int \frac{1}{x ^{2} + 4} d x

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

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

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

= \frac{1}{2} (x^{2} + 4 - 4 ln x^{2} + 4) + \frac{1}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2})

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} x^{2} + 2 - \frac{3}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

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