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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{2} ln tan (arcsec (x^{2})) + sec (arcsec (x^{2}))

Simplify

\frac{1}{2} ln tan (arcsec (x^{2})) + sec (arcsec (x^{2})) : \frac{1}{2} ln x^{4} - 1 + x^{2}

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x^{2} y - y^{2} + y^{3} x^{2})

\int 1 - 4 x d x

cot^{2} (sin (θ))

y = cos (sin (tan (π x)))

\frac{d y}{d x} = y^{(\frac{1}{3})}

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