What is the integral of (x^3)/(1-x^2) ?

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

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integral of (x^3)/(1-x^2)

\int \frac{x ^{3}}{1 - x ^{2}} d x

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

Solution steps

\int \frac{x ^{3}}{1 - x ^{2}} d x

Apply u-substitution

: \int - \frac{- u + 1}{2 u} d u

= \int - \frac{- u + 1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int \frac{- u + 1}{u} d u

Expand

\frac{- u + 1}{u} : - 1 + \frac{1}{u}

= - \frac{1}{2} \cdot \int - 1 + \frac{1}{u} d u

Apply the Sum Rule:

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

= - \frac{1}{2} (- \int 1 d u + \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= - \frac{1}{2} (- u + ln ∣ u ∣)

Substitute back

u = 1 - x^{2}

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

Simplify

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

Add a constant to the solution

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

x \to 4 - lim (2 - x - x^{2})

f (x) = (x^{3} + 9) e^{x}

\int_{e}^{e^{3}} \frac{15}{y} d y

\int x^{2} cos^{2} (x) d x

y^{^{''}} + 8 y^{^{'}} - 9 y = 0, y (1) = 2, y^{^{'}} (1) = 0

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