What is the integral of 1/(36x^2-1) ?

The integral of 1/(36x^2-1) is -1/12 (ln|6x+1|-ln|6x-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(36x^2-1)

\int \frac{1}{36 x ^{2} - 1} d x

- \frac{1}{12} (ln ∣ 6 x + 1 ∣ - ln ∣ 6 x - 1 ∣) + C

Solution steps

\int \frac{1}{36 x ^{2} - 1} d x

Factor

36 x^{2} - 1 : - (- 36 x^{2} + 1)

= \int \frac{1}{- ( - 36 x ^{2} + 1 )} d x

Take the constant out:

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

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

Apply Integral Substitution

= - \int \frac{1}{6 ( - 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}{6} \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}

= - \frac{1}{6} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = 6 x

= - \frac{1}{6} (\frac{ln ∣ 6 x + 1 ∣}{2} - \frac{ln ∣ 6 x - 1 ∣}{2})

Simplify

- \frac{1}{6} (\frac{ln ∣ 6 x + 1 ∣}{2} - \frac{ln ∣ 6 x - 1 ∣}{2}) : - \frac{1}{12} (ln ∣ 6 x + 1 ∣ - ln ∣ 6 x - 1 ∣)

= - \frac{1}{12} (ln ∣ 6 x + 1 ∣ - ln ∣ 6 x - 1 ∣)

Add a constant to the solution

= - \frac{1}{12} (ln ∣ 6 x + 1 ∣ - ln ∣ 6 x - 1 ∣) + C

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

\int \frac{11 x ^{2} + 5 x - 4}{2 x ^{3} - x ^{2}} d x

\int x^{2} x^{3} - 5 d x

\int \frac{x + 1}{x ^{4} + 3 x ^{2}} d x

\int \frac{x ^{2}}{x ^{2} + 4 x + 5} d x

What is the integral of 1/(36x^2-1) ?
The integral of 1/(36x^2-1) is -1/12 (ln|6x+1|-ln|6x-1|)+C