What is the integral of (6x^4+23x^2+4x+15)/(x^2+5) ?

The integral of (6x^4+23x^2+4x+15)/(x^2+5) is 2x^3-7x+10sqrt(5)arctan(x/(sqrt(5)))+2ln|x^2+5|+C

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integral of (6x^4+23x^2+4x+15)/(x^2+5)

\int \frac{6 x ^{4} + 23 x ^{2} + 4 x + 15}{x ^{2} + 5} d x

2 x^{3} - 7 x + 105 arctan (\frac{x}{5}) + 2 ln x^{2} + 5 + C

Solution steps

\int \frac{6 x ^{4} + 23 x ^{2} + 4 x + 15}{x ^{2} + 5} d x

Expand

\frac{6 x ^{4} + 23 x ^{2} + 4 x + 15}{x ^{2} + 5} : \frac{6 x ^{4}}{x ^{2} + 5} + \frac{23 x ^{2}}{x ^{2} + 5} + \frac{4 x}{x ^{2} + 5} + \frac{15}{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{6 x ^{4}}{x ^{2} + 5} d x + \int \frac{23 x ^{2}}{x ^{2} + 5} d x + \int \frac{4 x}{x ^{2} + 5} d x + \int \frac{15}{x ^{2} + 5} d x

\int \frac{6 x ^{4}}{x ^{2} + 5} d x = 2 x^{3} - 30 x + 305 arctan (\frac{x}{5})

\int \frac{23 x ^{2}}{x ^{2} + 5} d x = - 235 arctan (\frac{x}{5}) + 23 x

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

\int \frac{15}{x ^{2} + 5} d x = 35 arctan (\frac{x}{5})

= 2 x^{3} - 30 x + 305 arctan (\frac{x}{5}) - 235 arctan (\frac{x}{5}) + 23 x + 2 ln x^{2} + 5 + 35 arctan (\frac{x}{5})

Simplify

= 2 x^{3} - 7 x + 105 arctan (\frac{x}{5}) + 2 ln x^{2} + 5

Add a constant to the solution

= 2 x^{3} - 7 x + 105 arctan (\frac{x}{5}) + 2 ln x^{2} + 5 + C

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What is the integral of (6x^4+23x^2+4x+15)/(x^2+5) ?
The integral of (6x^4+23x^2+4x+15)/(x^2+5) is 2x^3-7x+10sqrt(5)arctan(x/(sqrt(5)))+2ln|x^2+5|+C