What is the integral of (x^2)/(sqrt(x^6-1)) ?

The integral of (x^2)/(sqrt(x^6-1)) is 1/3 ln|sqrt(x^6-1)+x^3|+C

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integral of (x^2)/(sqrt(x^6-1))

\int \frac{x ^{2}}{x ^{6} - 1} d x

\frac{1}{3} ln x^{6} - 1 + x^{3} + C

Solution steps

\int \frac{x ^{2}}{x ^{6} - 1} d x

Apply u-substitution

= \int \frac{1}{3 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}{3} \cdot \int \frac{1}{u ^{2} - 1} d u

Apply Trigonometric Substitution

= \frac{1}{3} \cdot \int sec (v) d v

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{3} ln tan (arcsec (x^{3})) + sec (arcsec (x^{3}))

Simplify

\frac{1}{3} ln tan (arcsec (x^{3})) + sec (arcsec (x^{3})) : \frac{1}{3} ln x^{6} - 1 + x^{3}

= \frac{1}{3} ln x^{6} - 1 + x^{3}

Add a constant to the solution

= \frac{1}{3} ln x^{6} - 1 + x^{3} + C

What is the integral of (x^2)/(sqrt(x^6-1)) ?
The integral of (x^2)/(sqrt(x^6-1)) is 1/3 ln|sqrt(x^6-1)+x^3|+C