What is the integral of (x^2)/(sqrt(x^6-9)) ?

The integral of (x^2)/(sqrt(x^6-9)) is 1/3 ln((|sqrt(x^6-9)+x^3|)/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2)/(sqrt(x^6-9))

\int \frac{x ^{2}}{x ^{6} - 9} d x

\frac{1}{3} ln (\frac{x ^{6} - 9 + x ^{3}}{3}) + C

Solution steps

\int \frac{x ^{2}}{x ^{6} - 9} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2} - 9} d u

= \int \frac{1}{3 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}{3} \cdot \int \frac{1}{u ^{2} - 9} 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}{3} x^{3})) + sec (arcsec (\frac{1}{3} x^{3}))

Simplify

\frac{1}{3} ln tan (arcsec (\frac{1}{3} x^{3})) + sec (arcsec (\frac{1}{3} x^{3})) : \frac{1}{3} ln (\frac{x ^{6} - 9 + x ^{3}}{3})

= \frac{1}{3} ln (\frac{x ^{6} - 9 + x ^{3}}{3})

Add a constant to the solution

= \frac{1}{3} ln (\frac{x ^{6} - 9 + x ^{3}}{3}) + C

(3, 0), (6, 2)

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

e^{x^{2}}

\frac{1}{s ^{2} + 16}

\frac{d y}{d x}, y = (sin (7 x))^{l n (x)}

What is the integral of (x^2)/(sqrt(x^6-9)) ?
The integral of (x^2)/(sqrt(x^6-9)) is 1/3 ln((|sqrt(x^6-9)+x^3|)/3)+C