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

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

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

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

ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - x + C

Solution steps

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

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

\int 2 t - 3 d t

\int (ln (x) + 1) d x

\frac{d}{d x} ((6 x)^{4 x})

\int \frac{1}{( 5 x - 2 ) ^{\frac{5}{2}}} d x

\frac{\partial}{\partial x} (\frac{2}{x ^{2} + y ^{2}})

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