What is the integral of (x^6)/(sqrt(x^{14)-5)} ?

The integral of (x^6)/(sqrt(x^{14)-5)} is 1/7 ln|sqrt(1/5 (x^{14)-5)}+1/(sqrt(5))x^7|+C

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integral of (x^6)/(sqrt(x^{14)-5)}

\int \frac{x ^{6}}{x ^{14} - 5} d x

\frac{1}{7} ln \frac{1}{5} (x^{14} - 5) + \frac{1}{5} x^{7} + C

Solution steps

\int \frac{x ^{6}}{x ^{14} - 5} d x

Apply u-substitution

: \int \frac{1}{7 u ^{2} - 5} d u

= \int \frac{1}{7 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}{7} \cdot \int \frac{1}{u ^{2} - 5} 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}{5} x^{7})) + sec (arcsec (\frac{1}{5} x^{7}))

Simplify

\frac{1}{7} ln tan (arcsec (\frac{1}{5} x^{7})) + sec (arcsec (\frac{1}{5} x^{7})) : \frac{1}{7} ln \frac{1}{5} (x^{14} - 5) + \frac{1}{5} x^{7}

= \frac{1}{7} ln \frac{1}{5} (x^{14} - 5) + \frac{1}{5} x^{7}

Add a constant to the solution

= \frac{1}{7} ln \frac{1}{5} (x^{14} - 5) + \frac{1}{5} x^{7} + C

f (x) = \frac{3 x + 5}{1 - x ^{2}}

f (x) = \frac{10 x}{x ^{2} - 6}

\int \frac{x ^{5}}{e ^{2 - 3 x^{6}}} d x

n = 1 \sum \infty \frac{1}{1 - 2 n}

3 t

What is the integral of (x^6)/(sqrt(x^{14)-5)} ?
The integral of (x^6)/(sqrt(x^{14)-5)} is 1/7 ln|sqrt(1/5 (x^{14)-5)}+1/(sqrt(5))x^7|+C