What is the integral of (x^2)/(sqrt(x^6-5)) ?

The integral of (x^2)/(sqrt(x^6-5)) is 1/3 ln|sqrt(1/5 (x^6-5))+1/(sqrt(5))x^3|+C

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integral of (x^2)/(sqrt(x^6-5))

\int \frac{x ^{2}}{x ^{6} - 5} d x

\frac{1}{3} ln \frac{1}{5} (x^{6} - 5) + \frac{1}{5} x^{3} + C

Solution steps

\int \frac{x ^{2}}{x ^{6} - 5} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2} - 5} d u

= \int \frac{1}{3 u ^{2} - 5} 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} - 5} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \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 (\frac{1}{5} x^{3})) + sec (arcsec (\frac{1}{5} x^{3}))

Simplify

\frac{1}{3} ln tan (arcsec (\frac{1}{5} x^{3})) + sec (arcsec (\frac{1}{5} x^{3})) : \frac{1}{3} ln \frac{1}{5} (x^{6} - 5) + \frac{1}{5} x^{3}

= \frac{1}{3} ln \frac{1}{5} (x^{6} - 5) + \frac{1}{5} x^{3}

Add a constant to the solution

= \frac{1}{3} ln \frac{1}{5} (x^{6} - 5) + \frac{1}{5} x^{3} + C

\frac{d}{d x} (\frac{x}{( x ^{2} - 1 ) ^{4}})

\int tan^{2} (x) \cdot sec (x) d x

\frac{1}{4} x^{2}, 3 - x, x = 1

\frac{d}{d x} (ln (ln (12 x)))

\int \frac{1}{2 sin ^{2} ( x ) + 3 cos ^{2} ( x )} d x

What is the integral of (x^2)/(sqrt(x^6-5)) ?
The integral of (x^2)/(sqrt(x^6-5)) is 1/3 ln|sqrt(1/5 (x^6-5))+1/(sqrt(5))x^3|+C