What is the integral of (40)/(16-x^2) ?

The integral of (40)/(16-x^2) is 10(1/2 ln| x/4+1|-1/2 ln| x/4-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (40)/(16-x^2)

\int \frac{40}{16 - x ^{2}} d x

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

Solution steps

\int \frac{40}{16 - x ^{2}} d x

Take the constant out:

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

= 40 \cdot \int \frac{1}{16 - x ^{2}} d x

Apply Integral Substitution

= 40 \cdot \int \frac{1}{4 ( - u ^{2} + 1 )} d u

Take the constant out:

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

= 40 \cdot \frac{1}{4} \cdot \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

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

= 40 \cdot \frac{1}{4} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{x}{4}

= 40 \cdot \frac{1}{4} (\frac{ln \frac{x}{4} + 1}{2} - \frac{ln \frac{x}{4} - 1}{2})

Simplify

40 \cdot \frac{1}{4} (\frac{ln \frac{x}{4} + 1}{2} - \frac{ln \frac{x}{4} - 1}{2}) : 10 (\frac{1}{2} ln \frac{x}{4} + 1 - \frac{1}{2} ln \frac{x}{4} - 1)

= 10 (\frac{1}{2} ln \frac{x}{4} + 1 - \frac{1}{2} ln \frac{x}{4} - 1)

Add a constant to the solution

= 10 (\frac{1}{2} ln \frac{x}{4} + 1 - \frac{1}{2} ln \frac{x}{4} - 1) + C

\frac{d ^{2}}{d x ^{2}} (\frac{8}{9 x ^{5}})

\frac{\partial}{\partial z} (\frac{x - y}{y + z})

\int (e^{t} + t^{e}) d t

\frac{2}{x}, 8 x, \frac{x}{2}

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

What is the integral of (40)/(16-x^2) ?
The integral of (40)/(16-x^2) is 10(1/2 ln| x/4+1|-1/2 ln| x/4-1|)+C