What is the integral of 6ln(x^2-1) ?

The integral of 6ln(x^2-1) is 6(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6ln(x^2-1)

\int 6 ln (x^{2} - 1) d x

6 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

Solution steps

\int 6 ln (x^{2} - 1) d x

Take the constant out:

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

= 6 \cdot \int ln (x^{2} - 1) d x

Apply Integration By Parts

: x ln (x^{2} - 1) - \int \frac{2 x ^{2}}{x ^{2} - 1} d x

= 6 (x ln (x^{2} - 1) - \int \frac{2 x ^{2}}{x ^{2} - 1} d x)

\int \frac{2 x ^{2}}{x ^{2} - 1} d x = - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x

= 6 (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x))

Simplify

= 6 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x)

Add a constant to the solution

= 6 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

\frac{d}{d t} (ln (t))

\frac{\partial}{\partial x} (e^{- j w n})

\frac{d y}{d x} = \frac{( x - y )}{( x + 2 y )}, y (0) = 1

\int (\frac{1}{1 + x + y}) d x

y^{^{''}} + 2 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the integral of 6ln(x^2-1) ?
The integral of 6ln(x^2-1) is 6(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C