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

The integral of ((x-3))/((x^2+4)) is 1/2 ln|x^2+4|-3/2 arctan(x/2)+C

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

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

\frac{1}{2} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2}) + C

Solution steps

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

Expand

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

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

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

= \frac{1}{2} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2}) + C

What is the integral of ((x-3))/((x^2+4)) ?
The integral of ((x-3))/((x^2+4)) is 1/2 ln|x^2+4|-3/2 arctan(x/2)+C