What is the integral of (x^4)/(sqrt(x^{10)-2)} ?

The integral of (x^4)/(sqrt(x^{10)-2)} is 1/5 ln|sqrt(1/2 (x^{10)-2)}+1/(sqrt(2))x^5|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^4)/(sqrt(x^{10)-2)}

\int \frac{x ^{4}}{x ^{10} - 2} d x

\frac{1}{5} ln \frac{1}{2} (x^{10} - 2) + \frac{1}{2} x^{5} + C

Solution steps

\int \frac{x ^{4}}{x ^{10} - 2} d x

Apply u-substitution

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

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

Simplify

\frac{1}{5} ln tan (arcsec (\frac{1}{2} x^{5})) + sec (arcsec (\frac{1}{2} x^{5})) : \frac{1}{5} ln \frac{1}{2} (x^{10} - 2) + \frac{1}{2} x^{5}

= \frac{1}{5} ln \frac{1}{2} (x^{10} - 2) + \frac{1}{2} x^{5}

Add a constant to the solution

= \frac{1}{5} ln \frac{1}{2} (x^{10} - 2) + \frac{1}{2} x^{5} + C

\frac{d}{d x} (y \cdot x^{2})

\int \frac{4 x ^{2}}{4 x - x ^{2}} d x

x \to 5 lim (x^{2} - 7 x + 6)

\frac{d}{d x} (8 x e^{2 x})

\int \frac{e ^{x}}{( 8 - e ^{x} ) ^{2}} d x

What is the integral of (x^4)/(sqrt(x^{10)-2)} ?
The integral of (x^4)/(sqrt(x^{10)-2)} is 1/5 ln|sqrt(1/2 (x^{10)-2)}+1/(sqrt(2))x^5|+C