What is the integral of (x+9)/(x^2+4) ?

The integral of (x+9)/(x^2+4) is 1/2 ln|x^2+4|+9/2 arctan(x/2)+C

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integral of (x+9)/(x^2+4)

\int \frac{x + 9}{x ^{2} + 4} d x

\frac{1}{2} ln x^{2} + 4 + \frac{9}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{x + 9}{x ^{2} + 4} d x

Expand

\frac{x + 9}{x ^{2} + 4} : \frac{x}{x ^{2} + 4} + \frac{9}{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}{x ^{2} + 4} d x + \int \frac{9}{x ^{2} + 4} d x

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

\int \frac{9}{x ^{2} + 4} d x = \frac{9}{2} arctan (\frac{x}{2})

= \frac{1}{2} ln x^{2} + 4 + \frac{9}{2} arctan (\frac{x}{2})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 + \frac{9}{2} arctan (\frac{x}{2}) + C

x \to \infty lim ((x!)^{2})

x^{2} - 12, 4 x

y = (x + 3)^{3}, at x = - \frac{1}{75}

\frac{d}{d x} (\frac{2}{( 2 - x ) ^{2}})

\frac{d}{d x} (\frac{2}{3} (x^{2} - 1)^{\frac{3}{2}})

What is the integral of (x+9)/(x^2+4) ?
The integral of (x+9)/(x^2+4) is 1/2 ln|x^2+4|+9/2 arctan(x/2)+C