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

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

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

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{u - 1}{u ^{2} + 1} d u

Expand

\frac{u - 1}{u ^{2} + 1} : \frac{u}{u ^{2} + 1} - \frac{1}{u ^{2} + 1}

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{u}{u ^{2} + 1} d u - \int \frac{1}{u ^{2} + 1} d u

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

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

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

Substitute back

u = \frac{x}{2}

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

(\frac{x}{2})^{2} = \frac{x ^{2}}{4}

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

Add a constant to the solution

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

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