What is the integral of (x^6)/(sqrt(x^{14)-9)} ?

The integral of (x^6)/(sqrt(x^{14)-9)} is 1/7 ln((|sqrt(x^{14)-9}+x^7|)/3)+C

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integral of (x^6)/(sqrt(x^{14)-9)}

\int \frac{x ^{6}}{x ^{14} - 9} d x

\frac{1}{7} ln (\frac{x ^{14} - 9 + x ^{7}}{3}) + C

Solution steps

\int \frac{x ^{6}}{x ^{14} - 9} d x

Apply u-substitution

: \int \frac{1}{7 u ^{2} - 9} d u

= \int \frac{1}{7 u ^{2} - 9} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{7} \cdot \int \frac{1}{u ^{2} - 9} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \frac{1}{7} \cdot \int sec (v) d v

Use the common integral:

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

= \frac{1}{7} ln ∣ tan (v) + sec (v) ∣

Substitute back

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

Simplify

\frac{1}{7} ln tan (arcsec (\frac{1}{3} x^{7})) + sec (arcsec (\frac{1}{3} x^{7})) : \frac{1}{7} ln (\frac{x ^{14} - 9 + x ^{7}}{3})

= \frac{1}{7} ln (\frac{x ^{14} - 9 + x ^{7}}{3})

Add a constant to the solution

= \frac{1}{7} ln (\frac{x ^{14} - 9 + x ^{7}}{3}) + C

4 + 3 x

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

x = y^{2} - 2, x = 6 - y^{2}

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

\frac{d}{d x} (arctan (e^{x} x^{2}))

What is the integral of (x^6)/(sqrt(x^{14)-9)} ?
The integral of (x^6)/(sqrt(x^{14)-9)} is 1/7 ln((|sqrt(x^{14)-9}+x^7|)/3)+C