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

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

Popular >

integral of (x^2+1)/(1-x^2)

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

ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - x + C

Solution steps

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

\frac{x ^{2} + 1}{1 - x ^{2}} = \frac{2}{1 - x ^{2}} - 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d x = x

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - x

Add a constant to the solution

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - x + C

x \to \frac{5 π}{6} lim (sin (x))

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

\int \frac{x ^{5}}{x ^{6} + 2} d x

x \to 9 lim (\frac{1}{( x - 9 ) ^{3}})

7 cos (7 x), 7 - 7 cos (7 x)

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