What is the integral of (x^3+4x-1)/(x^2+5) ?

The integral of (x^3+4x-1)/(x^2+5) is 1/2 x^2+5/2-1/2 ln|x^2+5|-1/(sqrt(5))arctan(x/(sqrt(5)))+C

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integral of (x^3+4x-1)/(x^2+5)

\int \frac{x ^{3} + 4 x - 1}{x ^{2} + 5} d x

\frac{1}{2} x^{2} + \frac{5}{2} - \frac{1}{2} ln x^{2} + 5 - \frac{1}{5} arctan (\frac{x}{5}) + C

Solution steps

\int \frac{x ^{3} + 4 x - 1}{x ^{2} + 5} d x

Expand

\frac{x ^{3} + 4 x - 1}{x ^{2} + 5} : \frac{x ^{3}}{x ^{2} + 5} + \frac{4 x}{x ^{2} + 5} - \frac{1}{x ^{2} + 5}

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} + 5} d x + \int \frac{4 x}{x ^{2} + 5} d x - \int \frac{1}{x ^{2} + 5} d x

\int \frac{x ^{3}}{x ^{2} + 5} d x = \frac{1}{2} (x^{2} + 5 - 5 ln x^{2} + 5)

\int \frac{4 x}{x ^{2} + 5} d x = 2 ln x^{2} + 5

\int \frac{1}{x ^{2} + 5} d x = \frac{1}{5} arctan (\frac{x}{5})

= \frac{1}{2} (x^{2} + 5 - 5 ln x^{2} + 5) + 2 ln x^{2} + 5 - \frac{1}{5} arctan (\frac{x}{5})

Simplify

\frac{1}{2} (x^{2} + 5 - 5 ln x^{2} + 5) + 2 ln x^{2} + 5 - \frac{1}{5} arctan (\frac{x}{5}) : \frac{1}{2} x^{2} + \frac{5}{2} - \frac{1}{2} ln x^{2} + 5 - \frac{1}{5} arctan (\frac{x}{5})

= \frac{1}{2} x^{2} + \frac{5}{2} - \frac{1}{2} ln x^{2} + 5 - \frac{1}{5} arctan (\frac{x}{5})

Add a constant to the solution

= \frac{1}{2} x^{2} + \frac{5}{2} - \frac{1}{2} ln x^{2} + 5 - \frac{1}{5} arctan (\frac{x}{5}) + C

What is the integral of (x^3+4x-1)/(x^2+5) ?
The integral of (x^3+4x-1)/(x^2+5) is 1/2 x^2+5/2-1/2 ln|x^2+5|-1/(sqrt(5))arctan(x/(sqrt(5)))+C