What is the integral of 8/(4+x^2) ?

The integral of 8/(4+x^2) is 4arctan(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 8/(4+x^2)

\int \frac{8}{4 + x ^{2}} d x

4 arctan (\frac{x}{2}) + C

Solution steps

\int \frac{8}{4 + x ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{4 + x ^{2}} d x

Apply Integral Substitution

= 8 \cdot \int \frac{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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 8 \cdot \frac{1}{2} arctan (u)

Substitute back

u = \frac{x}{2}

= 8 \cdot \frac{1}{2} arctan (\frac{x}{2})

Simplify

8 \cdot \frac{1}{2} arctan (\frac{x}{2}) : 4 arctan (\frac{x}{2})

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

Add a constant to the solution

= 4 arctan (\frac{x}{2}) + C

\frac{\partial}{\partial x} (x^{3} e^{x^{2} + y} - 12 y^{2})

y = - 2 x sin (x^{2})

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

x \to 10 lim (\frac{10 x}{2 x + 5})

\int - 45 t^{\frac{2}{3}} - 80 d t