What is the integral of 1/(x*sqrt(9-4x^2)) ?

The integral of 1/(x*sqrt(9-4x^2)) is -1/6 (ln|(sqrt(9-4x^2))/3+1|-ln|(sqrt(9-4x^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(9-4x^2))

\int \frac{1}{x \cdot 9 - 4 x ^{2}} d x

- \frac{1}{6} (ln \frac{9 - 4 x ^{2}}{3} + 1 - ln \frac{9 - 4 x ^{2}}{3} - 1) + C

Solution steps

\int \frac{1}{x 9 - 4 x ^{2}} d x

Apply u-substitution

: \int \frac{1}{2 u 9 - 4 u} d u

= \int \frac{1}{2 u 9 - 4 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u 9 - 4 u} d u

Apply u-substitution

: \int \frac{2}{v ^{2} - 9} d v

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

Take the constant out:

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

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

Apply Integral Substitution

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{3 ( w ^{2} - 1 )} d w

Take the constant out:

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

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

Factor

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} - \frac{ln \frac{9 - 4 x ^{2}}{3} + 1}{2} - \frac{ln \frac{9 - 4 x ^{2}}{3} - 1}{2}

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{1}{3} - \frac{ln \frac{9 - 4 x ^{2}}{3} + 1}{2} - \frac{ln \frac{9 - 4 x ^{2}}{3} - 1}{2} : - \frac{1}{6} (ln \frac{9 - 4 x ^{2}}{3} + 1 - ln \frac{9 - 4 x ^{2}}{3} - 1)

= - \frac{1}{6} (ln \frac{9 - 4 x ^{2}}{3} + 1 - ln \frac{9 - 4 x ^{2}}{3} - 1)

Add a constant to the solution

= - \frac{1}{6} (ln \frac{9 - 4 x ^{2}}{3} + 1 - ln \frac{9 - 4 x ^{2}}{3} - 1) + C

(sin (5 t))^{^{'}}

\int \frac{1}{36 e ^{- 3 x} + e ^{3 x}} d x

\frac{d}{d x} (tan (x) - sin (x))

h (x) = - 6 x^{- 2}

\frac{2}{( 1 - x ) ^{3}}

What is the integral of 1/(x*sqrt(9-4x^2)) ?
The integral of 1/(x*sqrt(9-4x^2)) is -1/6 (ln|(sqrt(9-4x^2))/3+1|-ln|(sqrt(9-4x^2))/3-1|)+C