What is the integral of 1/(16-9x^2) ?

The integral of 1/(16-9x^2) is 1/24 (ln| 3/4 x+1|-ln| 3/4 x-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/(16-9x^2)

\int \frac{1}{16 - 9 x ^{2}} d x

\frac{1}{24} (ln \frac{3}{4} x + 1 - ln \frac{3}{4} x - 1) + C

Solution steps

\int \frac{1}{16 - 9 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{12 ( - 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}{12} \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}{12} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{3}{4} x

= \frac{1}{12} (\frac{ln \frac{3}{4} x + 1}{2} - \frac{ln \frac{3}{4} x - 1}{2})

Simplify

\frac{1}{12} (\frac{ln \frac{3}{4} x + 1}{2} - \frac{ln \frac{3}{4} x - 1}{2}) : \frac{1}{24} (ln \frac{3}{4} x + 1 - ln \frac{3}{4} x - 1)

= \frac{1}{24} (ln \frac{3}{4} x + 1 - ln \frac{3}{4} x - 1)

Add a constant to the solution

= \frac{1}{24} (ln \frac{3}{4} x + 1 - ln \frac{3}{4} x - 1) + C

\int t^{4} + t^{2} d t

\int 15 x - x d x

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

\int csc^{10} (x) cot^{3} (x) d x

\int \frac{ln ( x ^{2} + x + 1 )}{x ^{3}} d x

What is the integral of 1/(16-9x^2) ?
The integral of 1/(16-9x^2) is 1/24 (ln| 3/4 x+1|-ln| 3/4 x-1|)+C