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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

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

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

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

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

(\frac{x}{4})^{2} = \frac{x ^{2}}{16}

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{x ^{25}})

(3, - 5), (- 1, 3)

sin^{7} (cos (5 x))

f (x) = 3 x^{2} + 2 x, at x = - 3

\frac{\partial}{\partial y} ((e^{x} - x) cos (y))

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