What is the integral of 1/(3x^2-5) ?

The integral of 1/(3x^2-5) is -1/(2sqrt(15))(ln|sqrt(3/5)x+1|-ln|sqrt(3/5)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/(3x^2-5)

\int \frac{1}{3 x ^{2} - 5} d x

- \frac{1}{2 15} (ln \frac{3}{5} x + 1 - ln \frac{3}{5} x - 1) + C

Solution steps

\int \frac{1}{3 x ^{2} - 5} d x

Apply Integral Substitution

= \int \frac{1}{15 ( 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}{15} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{15} \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

= \frac{1}{15} (- \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}{15} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{3}{5} x

= \frac{1}{15} - \frac{ln \frac{3}{5} x + 1}{2} - \frac{ln \frac{3}{5} x - 1}{2}

Simplify

\frac{1}{15} - \frac{ln \frac{3}{5} x + 1}{2} - \frac{ln \frac{3}{5} x - 1}{2} : - \frac{1}{2 15} (ln \frac{3}{5} x + 1 - ln \frac{3}{5} x - 1)

= - \frac{1}{2 15} (ln \frac{3}{5} x + 1 - ln \frac{3}{5} x - 1)

Add a constant to the solution

= - \frac{1}{2 15} (ln \frac{3}{5} x + 1 - ln \frac{3}{5} x - 1) + C

\int \frac{1}{cos ^{5} ( x ) sin ( x )} d x

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

\int (\frac{3}{91}) x^{2} d x

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

\int \frac{x}{( x ^{2} - 5 ) ^{\frac{1}{7}}} d x

What is the integral of 1/(3x^2-5) ?
The integral of 1/(3x^2-5) is -1/(2sqrt(15))(ln|sqrt(3/5)x+1|-ln|sqrt(3/5)x-1|)+C