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

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

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integral of x/(sqrt(x^4-16))

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{2 u ^{2} - 16} 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} - 16} 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 (\frac{1}{4} x^{2})) + sec (arcsec (\frac{1}{4} x^{2}))

Simplify

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

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

Add a constant to the solution

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

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