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

The integral of 1/(x*sqrt(9-x^2)) is -1/6 (ln|(sqrt(9-x^2))/3+1|-ln|(sqrt(9-x^2))/3-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x*sqrt(9-x^2))

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

- \frac{1}{6} (ln \frac{9 - x ^{2}}{3} + 1 - ln \frac{9 - x ^{2}}{3} - 1) + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{- u ^{2} + 9} d u

= \int - \frac{1}{- u ^{2} + 9} d u

Take the constant out:

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

= - \int \frac{1}{- u ^{2} + 9} d u

Apply Integral Substitution

= - \int \frac{1}{3 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= - \frac{1}{3} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

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

= - \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= - \frac{1}{3} \frac{ln \frac{9 - x ^{2}}{3} + 1}{2} - \frac{ln \frac{9 - x ^{2}}{3} - 1}{2}

Simplify

- \frac{1}{3} \frac{ln \frac{9 - x ^{2}}{3} + 1}{2} - \frac{ln \frac{9 - x ^{2}}{3} - 1}{2} : - \frac{1}{6} (ln \frac{9 - x ^{2}}{3} + 1 - ln \frac{9 - x ^{2}}{3} - 1)

= - \frac{1}{6} (ln \frac{9 - x ^{2}}{3} + 1 - ln \frac{9 - x ^{2}}{3} - 1)

Add a constant to the solution

= - \frac{1}{6} (ln \frac{9 - x ^{2}}{3} + 1 - ln \frac{9 - x ^{2}}{3} - 1) + C

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

\frac{\partial}{\partial x} (tan (3 x - 5 y))

\frac{\partial}{\partial x} (y^{2} - 1)

\int \frac{x}{x - 1} d x

n = 0 \sum \infty (\frac{7}{18})^{n}

What is the integral of 1/(x*sqrt(9-x^2)) ?
The integral of 1/(x*sqrt(9-x^2)) is -1/6 (ln|(sqrt(9-x^2))/3+1|-ln|(sqrt(9-x^2))/3-1|)+C