What is the integral of (5x)/(sqrt(x^4-1)) ?

The integral of (5x)/(sqrt(x^4-1)) is 5/2 ln|sqrt(x^4-1)+x^2|+C

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integral of (5x)/(sqrt(x^4-1))

\int \frac{5 x}{x ^{4} - 1} d x

\frac{5}{2} ln x^{4} - 1 + x^{2} + C

Solution steps

\int \frac{5 x}{x ^{4} - 1} d x

Take the constant out:

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

= 5 \cdot \int \frac{x}{x ^{4} - 1} d x

Apply u-substitution

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

= 5 \cdot \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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} - 1} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= 5 \cdot \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

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

= 5 \cdot \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 5 \cdot \frac{1}{2} ln tan (arcsec (x^{2})) + sec (arcsec (x^{2}))

Simplify

5 \cdot \frac{1}{2} ln tan (arcsec (x^{2})) + sec (arcsec (x^{2})) : \frac{5}{2} ln x^{4} - 1 + x^{2}

= \frac{5}{2} ln x^{4} - 1 + x^{2}

Add a constant to the solution

= \frac{5}{2} ln x^{4} - 1 + x^{2} + C

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What is the integral of (5x)/(sqrt(x^4-1)) ?
The integral of (5x)/(sqrt(x^4-1)) is 5/2 ln|sqrt(x^4-1)+x^2|+C