What is the integral of x/(3sqrt(x^2-8)) ?

The integral of x/(3sqrt(x^2-8)) is 1/3 sqrt(x^2-8)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/(3sqrt(x^2-8))

\int \frac{x}{3 x ^{2} - 8} d x

\frac{1}{3} x^{2} - 8 + C

Solution steps

\int \frac{x}{3 x ^{2} - 8} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{x}{x ^{2} - 8} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= \frac{1}{3} \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{3} \cdot \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} - 8

= \frac{1}{3} \cdot \frac{1}{2} \cdot 2 (x^{2} - 8)^{\frac{1}{2}}

Simplify

\frac{1}{3} \cdot \frac{1}{2} \cdot 2 (x^{2} - 8)^{\frac{1}{2}} : \frac{1}{3} x^{2} - 8

= \frac{1}{3} x^{2} - 8

Add a constant to the solution

= \frac{1}{3} x^{2} - 8 + C

\int \frac{1}{4 - 16 x ^{2}} d x

y = e, y = e^{x}, y = e^{- x}

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 14

\int sin (X) d X

x \to 2 + lim (\frac{- 1}{x ^{2} - 4})

What is the integral of x/(3sqrt(x^2-8)) ?
The integral of x/(3sqrt(x^2-8)) is 1/3 sqrt(x^2-8)+C