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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x)/(4-x^4)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \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}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

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

x \to \infty lim ((x)^{\frac{l n ( 2 )}{1 + l n ( x )}})

f (x) = \frac{100}{( 1 + 5 x ) ^{3}}

\int x^{3} + sin (x) + cos (x) d x

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