What is the integral of (6x^4+23x^2+4x+15)/(x^2+3) ?

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

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integral of (6x^4+23x^2+4x+15)/(x^2+3)

\int \frac{6 x ^{4} + 23 x ^{2} + 4 x + 15}{x ^{2} + 3} d x

2 x^{3} + 5 x + 2 ln x^{2} + 3 + C

Solution steps

\int \frac{6 x ^{4} + 23 x ^{2} + 4 x + 15}{x ^{2} + 3} d x

Expand

\frac{6 x ^{4} + 23 x ^{2} + 4 x + 15}{x ^{2} + 3} : \frac{6 x ^{4}}{x ^{2} + 3} + \frac{23 x ^{2}}{x ^{2} + 3} + \frac{4 x}{x ^{2} + 3} + \frac{15}{x ^{2} + 3}

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} + 3} d x + \int \frac{23 x ^{2}}{x ^{2} + 3} d x + \int \frac{4 x}{x ^{2} + 3} d x + \int \frac{15}{x ^{2} + 3} d x

\int \frac{6 x ^{4}}{x ^{2} + 3} d x = 2 x^{3} - 18 x + 183 arctan (\frac{x}{3})

\int \frac{23 x ^{2}}{x ^{2} + 3} d x = - 233 arctan (\frac{x}{3}) + 23 x

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

\int \frac{15}{x ^{2} + 3} d x = 53 arctan (\frac{x}{3})

= 2 x^{3} - 18 x + 183 arctan (\frac{x}{3}) - 233 arctan (\frac{x}{3}) + 23 x + 2 ln x^{2} + 3 + 53 arctan (\frac{x}{3})

Simplify

= 2 x^{3} + 5 x + 2 ln x^{2} + 3

Add a constant to the solution

= 2 x^{3} + 5 x + 2 ln x^{2} + 3 + C

\frac{d}{d x} (x^{2} sin^{7} (x) + x cos^{- 2} (x))

4 y^{^{''}} + 2 y^{^{'}} + 2 y = 0

y = cos (x), at x = \frac{π}{2}

\frac{d}{d x} (x + x x)

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

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