What is the integral of (4x^3-x^2+16x)/(x^2+4) ?

The integral of (4x^3-x^2+16x)/(x^2+4) is 2x^2-x+2arctan(x/2)+8+C

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integral of (4x^3-x^2+16x)/(x^2+4)

\int \frac{4 x ^{3} - x ^{2} + 16 x}{x ^{2} + 4} d x

2 x^{2} - x + 2 arctan (\frac{x}{2}) + 8 + C

Solution steps

\int \frac{4 x ^{3} - x ^{2} + 16 x}{x ^{2} + 4} d x

Expand

\frac{4 x ^{3} - x ^{2} + 16 x}{x ^{2} + 4} : \frac{4 x ^{3}}{x ^{2} + 4} - \frac{x ^{2}}{x ^{2} + 4} + \frac{16 x}{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{4 x ^{3}}{x ^{2} + 4} d x - \int \frac{x ^{2}}{x ^{2} + 4} d x + \int \frac{16 x}{x ^{2} + 4} d x

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

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

\int \frac{16 x}{x ^{2} + 4} d x = 8 ln x^{2} + 4

= 2 (x^{2} + 4 - 4 ln x^{2} + 4) - (- 2 arctan (\frac{x}{2}) + x) + 8 ln x^{2} + 4

Simplify

2 (x^{2} + 4 - 4 ln x^{2} + 4) - (- 2 arctan (\frac{x}{2}) + x) + 8 ln x^{2} + 4 : 2 x^{2} - x + 2 arctan (\frac{x}{2}) + 8

= 2 x^{2} - x + 2 arctan (\frac{x}{2}) + 8

Add a constant to the solution

= 2 x^{2} - x + 2 arctan (\frac{x}{2}) + 8 + C

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What is the integral of (4x^3-x^2+16x)/(x^2+4) ?
The integral of (4x^3-x^2+16x)/(x^2+4) is 2x^2-x+2arctan(x/2)+8+C