What is the integral of (x^4)/(sqrt(x^{10)-1)} ?

The integral of (x^4)/(sqrt(x^{10)-1)} is 1/5 ln|sqrt(x^{10)-1}+x^5|+C

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integral of (x^4)/(sqrt(x^{10)-1)}

\int \frac{x ^{4}}{x ^{10} - 1} d x

\frac{1}{5} ln x^{10} - 1 + x^{5} + C

Solution steps

\int \frac{x ^{4}}{x ^{10} - 1} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \frac{1}{5} \cdot \int sec (v) d v

Use the common integral:

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

= \frac{1}{5} ln ∣ tan (v) + sec (v) ∣

Substitute back

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

Simplify

\frac{1}{5} ln tan (arcsec (x^{5})) + sec (arcsec (x^{5})) : \frac{1}{5} ln x^{10} - 1 + x^{5}

= \frac{1}{5} ln x^{10} - 1 + x^{5}

Add a constant to the solution

= \frac{1}{5} ln x^{10} - 1 + x^{5} + C

f (x) = \frac{7}{x - 6}

e^{k x}

\frac{\partial}{\partial x} (2 x^{2} + 4 y + 1)

f (x) = \frac{1}{2} ln (x)

\int \frac{3 x - x ^{3} + 1}{x ^{4}} d x

What is the integral of (x^4)/(sqrt(x^{10)-1)} ?
The integral of (x^4)/(sqrt(x^{10)-1)} is 1/5 ln|sqrt(x^{10)-1}+x^5|+C