What is the integral of 5/(x^2-16) ?

The integral of 5/(x^2-16) is -5/8 (ln| x/4+1|-ln| x/4-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 5/(x^2-16)

\int \frac{5}{x ^{2} - 16} d x

- \frac{5}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + C

Solution steps

\int \frac{5}{x ^{2} - 16} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{2} - 16} d x

Apply Integral Substitution

= 5 \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

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{4} (- \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}

= 5 \cdot \frac{1}{4} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{x}{4}

= 5 \cdot \frac{1}{4} (- (\frac{ln \frac{x}{4} + 1}{2} - \frac{ln \frac{x}{4} - 1}{2}))

Simplify

5 \cdot \frac{1}{4} (- (\frac{ln \frac{x}{4} + 1}{2} - \frac{ln \frac{x}{4} - 1}{2})) : - \frac{5}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1)

= - \frac{5}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1)

Add a constant to the solution

= - \frac{5}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + C

\int 10 x (1 - x)^{5} d x

\int x^{2} (7 x^{3} - 3)^{5} d x

\int (x^{3} + 2 x) (x^{4} + 4 x^{2} + 4)^{7} d x

\int \frac{1 - cos ( 2 u )}{2} d u

\int \frac{( - 5 x + 15 )}{( x ^{2} + 2 x + 4 ) ^{2}} d x

What is the integral of 5/(x^2-16) ?
The integral of 5/(x^2-16) is -5/8 (ln| x/4+1|-ln| x/4-1|)+C