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

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

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integral of ln(1-x^2)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

x \to - 7 + lim (x + 7)

\frac{\partial}{\partial x} (12 x^{2} - 9 x y + 16 y^{2})

\int cos (t) \cdot sin (t) d t

\int_{- 1}^{1} 1 5^{x} d x

y^{^{''}} + 441 y = sec (21 x)

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