What is the integral of sqrt(1+(4x^2)/((1-x^2)^2)) ?

The integral of sqrt(1+(4x^2)/((1-x^2)^2)) is ln|x+1|-ln|x-1|-x+C

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integral of sqrt(1+(4x^2)/((1-x^2)^2))

\int 1 + \frac{4 x ^{2}}{( 1 - x ^{2} ) ^{2}} d x

ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - x + C

Solution steps

\int 1 + \frac{4 x ^{2}}{( 1 - x ^{2} ) ^{2}} d x

Simplify

1 + \frac{4 x ^{2}}{( 1 - x ^{2} ) ^{2}} : \frac{x ^{2} + 1}{1 - x ^{2}}

= \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

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = - 7

\frac{\partial}{\partial y} (3.25 yz cos (x yz))

\frac{d}{d x} (x^{\frac{1}{3}} (x^{2} - 4))

5 + 4 e - 2 t

\frac{\partial}{\partial x} (x^{3} y + x y^{2} + x y + x)

What is the integral of sqrt(1+(4x^2)/((1-x^2)^2)) ?
The integral of sqrt(1+(4x^2)/((1-x^2)^2)) is ln|x+1|-ln|x-1|-x+C