What is the integral of-7ln(x^2-1) ?

The integral of-7ln(x^2-1) is -7(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-7ln(x^2-1)

\int - 7 ln (x^{2} - 1) d x

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

Solution steps

\int - 7 ln (x^{2} - 1) d x

Take the constant out:

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

= - 7 \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

= - 7 (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

= - 7 (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x))

Simplify

= - 7 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x)

Add a constant to the solution

= - 7 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

n = 2 \sum \infty \frac{1}{n ( n - 1 )}

\frac{d y}{d x}, 4 y^{2} - 5 y + 8 = 2 x^{2} - 3 x

\int 2 x (1 + x^{2})^{2} d x

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{2 x}

\frac{d}{d x} (5 e^{- x} + e^{3 x})

What is the integral of-7ln(x^2-1) ?
The integral of-7ln(x^2-1) is -7(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C