What is the integral of (x^2)/(sqrt(x^6-3)) ?

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

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integral of (x^2)/(sqrt(x^6-3))

\int \frac{x ^{2}}{x ^{6} - 3} d x

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

Solution steps

\int \frac{x ^{2}}{x ^{6} - 3} d x

Apply u-substitution

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

= \int \frac{1}{3 u ^{2} - 3} 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} - 3} 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}{3} x^{3})) + sec (arcsec (\frac{1}{3} x^{3}))

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} y = arccos (8 x^{2})

- \frac{1}{1 - x}

\frac{d}{d x} (4 + 2 x)

\int \frac{20}{1 + t ^{2}} d t

\frac{\partial}{\partial x} (2 e^{y} x y)

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