What is the integral of-10ln(x^2-1) ?

The integral of-10ln(x^2-1) is -10(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-10ln(x^2-1)

\int - 10 ln (x^{2} - 1) d x

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

Solution steps

\int - 10 ln (x^{2} - 1) d x

Take the constant out:

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

= - 10 \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

= - 10 (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

= - 10 (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x))

Simplify

= - 10 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x)

Add a constant to the solution

= - 10 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

\frac{d}{d x} (cos (π x))

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

\int \frac{1}{36 - 6 t ^{2}} d t

f (x) = 2 - x^{2}

What is the integral of-10ln(x^2-1) ?
The integral of-10ln(x^2-1) is -10(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C