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

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

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integral of x/(16-x^4)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x - x y + 3)

\int_{4}^{9} \frac{t - 3}{t} d t

x \to 0 lim (cos^{2} (5 x))

x \to - \infty lim (6)

\int (\frac{1}{x}) d x

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