What is the integral of (4x^2)/(4-9x^6) ?

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

\int \frac{4 x ^{2}}{4 - 9 x ^{6}} d x

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

Solution steps

\int \frac{4 x ^{2}}{4 - 9 x ^{6}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2}}{4 - 9 x ^{6}} d x

Apply u-substitution

: \int \frac{1}{3 ( 4 - 9 u ^{2} )} d u

= 4 \cdot \int \frac{1}{3 ( 4 - 9 u ^{2} )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{4 - 9 u ^{2}} d u

Apply Integral Substitution

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{6 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \frac{1}{6} \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}

= 4 \cdot \frac{1}{3} \cdot \frac{1}{6} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

\int 6 x^{3} - 3 x d x

\int \frac{1}{9 + 5 x ^{2}} d x

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

\int (10 x^{4} + 2 x + 1) d x

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