What is the integral of (x^2)/(sqrt(x^6-7)) ?

The integral of (x^2)/(sqrt(x^6-7)) is 1/3 ln|sqrt(1/7 (x^6-7))+1/(sqrt(7))x^3|+C

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integral of (x^2)/(sqrt(x^6-7))

\int \frac{x ^{2}}{x ^{6} - 7} d x

\frac{1}{3} ln \frac{1}{7} (x^{6} - 7) + \frac{1}{7} x^{3} + C

Solution steps

\int \frac{x ^{2}}{x ^{6} - 7} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2} - 7} d u

= \int \frac{1}{3 u ^{2} - 7} 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} - 7} 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}{7} x^{3})) + sec (arcsec (\frac{1}{7} x^{3}))

Simplify

\frac{1}{3} ln tan (arcsec (\frac{1}{7} x^{3})) + sec (arcsec (\frac{1}{7} x^{3})) : \frac{1}{3} ln \frac{1}{7} (x^{6} - 7) + \frac{1}{7} x^{3}

= \frac{1}{3} ln \frac{1}{7} (x^{6} - 7) + \frac{1}{7} x^{3}

Add a constant to the solution

= \frac{1}{3} ln \frac{1}{7} (x^{6} - 7) + \frac{1}{7} x^{3} + C

\frac{d}{d y} (2 x^{3} - x^{2} y)

\int \frac{1}{x ^{2} - 12 x + 35} d x

3 x^{^{'}} - 5 x = 12

\frac{d y}{d x} = \frac{1}{2} x + y - 1

y^{^{''''}} + 18 y^{^{''}} + 81 y = 0

What is the integral of (x^2)/(sqrt(x^6-7)) ?
The integral of (x^2)/(sqrt(x^6-7)) is 1/3 ln|sqrt(1/7 (x^6-7))+1/(sqrt(7))x^3|+C