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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply Integral Substitution

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

Expand

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

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

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

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

Substitute back

u = \frac{x}{2}

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

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

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

Add a constant to the solution

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

\int 225 - 25 x^{2} d x

\int sec (x) cos^{2} (x) d x

\int 2 sin (3 x + 1) d x

\int \frac{1}{36 x ^{2} + 100} d x

\int 27 tan^{6} (x) d x

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