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

The integral of (x^4)/(1-x^2) is -(x^3)/3-x+(ln|x+1|)/2-(ln|x-1|)/2+C

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

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

- \frac{x ^{3}}{3} - x + \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} + C

Solution steps

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

Long division

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

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

\int x^{2} d x = \frac{x ^{3}}{3}

\int 1 d x = x

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

= - \frac{x ^{3}}{3} - x + \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}

Add a constant to the solution

= - \frac{x ^{3}}{3} - x + \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} + C

\int x^{2} x^{3} d x

\int 7 v (v^{2} + 6)^{2} d v

6 x - x^{3}, ∣ 2 \cdot x ∣

ln (16 + x^{2})

v \to 1 lim (\frac{v ^{2} - 5}{v})

What is the integral of (x^4)/(1-x^2) ?
The integral of (x^4)/(1-x^2) is -(x^3)/3-x+(ln|x+1|)/2-(ln|x-1|)/2+C