What is the integral of ((x))/(16x^4-1) ?

The integral of ((x))/(16x^4-1) is -1/16 (ln|4x^2+1|-ln|4x^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 ((x))/(16x^4-1)

\int \frac{( x )}{16 x ^{4} - 1} d x

- \frac{1}{16} (ln 4 x^{2} + 1 - ln 4 x^{2} - 1) + C

Solution steps

\int \frac{x}{16 x ^{4} - 1} d x

Apply u-substitution

: \int \frac{1}{2 ( 16 u ^{2} - 1 )} d u

= \int \frac{1}{2 ( 16 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}{2} \cdot \int \frac{1}{16 u ^{2} - 1} d u

Factor

16 u^{2} - 1 : - (- 16 u^{2} + 1)

= \frac{1}{2} \cdot \int \frac{1}{- ( - 16 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}{2} (- \int \frac{1}{- 16 u ^{2} + 1} d u)

Apply Integral Substitution

= \frac{1}{2} (- \int \frac{1}{4 ( - 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}{2} (- \frac{1}{4} \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}{2} (- \frac{1}{4} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

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

Simplify

\frac{1}{2} (- \frac{1}{4} (\frac{ln 4 x ^{2} + 1}{2} - \frac{ln 4 x ^{2} - 1}{2})) : - \frac{1}{16} (ln 4 x^{2} + 1 - ln 4 x^{2} - 1)

= - \frac{1}{16} (ln 4 x^{2} + 1 - ln 4 x^{2} - 1)

Add a constant to the solution

= - \frac{1}{16} (ln 4 x^{2} + 1 - ln 4 x^{2} - 1) + C

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

x \to 7 - lim (\frac{3}{x - 7})

\int x^{2} + x + 3 d x

x \to 0 + lim (\frac{x ^{2}}{6} - \frac{7}{x})

\int 6 cos^{2} (x t a) n^{3} x d x

What is the integral of ((x))/(16x^4-1) ?
The integral of ((x))/(16x^4-1) is -1/16 (ln|4x^2+1|-ln|4x^2-1|)+C