What is the integral of (x-1)/(x^2+4) ?

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

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integral of (x-1)/(x^2+4)

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

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

Solution steps

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

Expand

\frac{x - 1}{x ^{2} + 4} : \frac{x}{x ^{2} + 4} - \frac{1}{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{1}{x ^{2} + 4} d x

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

\int \frac{1}{x ^{2} + 4} d x = \frac{1}{2} arctan (\frac{x}{2})

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

Add a constant to the solution

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

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

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

\frac{1 - 4 t}{5 + t}

\frac{3 s}{s ^{2} + 4}

x + 6 y = 8, x + 8 = y^{2}

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