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

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

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integral of 1/(x(x^4-1))

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 ( u + 1 ) u} d u

= \int \frac{1}{4 ( u + 1 ) u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{( u + 1 ) u} d u

Take the partial fraction of

\frac{1}{( u + 1 ) u} : - \frac{1}{u + 1} + \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (- \int \frac{1}{u + 1} d u + \int \frac{1}{u} d u)

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{4} (- ln ∣ u + 1 ∣ + ln ∣ u ∣)

Substitute back

u = - 1 + x^{4}

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

Simplify

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

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

Add a constant to the solution

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

(2 x y - 9 x^{2}) d x + (2 y + x^{2} + 1) d y = 0

\int \frac{4 x}{x ^{2} + 2 x - 3} d x

f (x) = tan (x), at x = 0

y^{^{'''}} + y^{^{''}} - 8 y^{^{'}} - 12 y = 0

(t (x) - t (x)^{3}) v^{'} + 2 v = 0

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