What is the integral of (x+9)/(x^2+16) ?

The integral of (x+9)/(x^2+16) is 1/2 ln|x^2+16|+9/4 arctan(x/4)+C

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integral of (x+9)/(x^2+16)

\int \frac{x + 9}{x ^{2} + 16} d x

\frac{1}{2} ln x^{2} + 16 + \frac{9}{4} arctan (\frac{x}{4}) + C

Solution steps

\int \frac{x + 9}{x ^{2} + 16} d x

Expand

\frac{x + 9}{x ^{2} + 16} : \frac{x}{x ^{2} + 16} + \frac{9}{x ^{2} + 16}

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} + 16} d x + \int \frac{9}{x ^{2} + 16} d x

\int \frac{x}{x ^{2} + 16} d x = \frac{1}{2} ln x^{2} + 16

\int \frac{9}{x ^{2} + 16} d x = \frac{9}{4} arctan (\frac{x}{4})

= \frac{1}{2} ln x^{2} + 16 + \frac{9}{4} arctan (\frac{x}{4})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 16 + \frac{9}{4} arctan (\frac{x}{4}) + C

\frac{d}{d x} (9 x^{2} - 2)

\int \frac{6 x ^{2} - 4 x - 25}{x - 2} d x

h \to 0 lim (\frac{( 1 + h ) - 1}{h})

x \to - 2 lim (x^{2} - 3)

\frac{d}{d x} ([x + (x + sin^{2} (x))^{7}]^{3})

What is the integral of (x+9)/(x^2+16) ?
The integral of (x+9)/(x^2+16) is 1/2 ln|x^2+16|+9/4 arctan(x/4)+C