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

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

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

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

ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1 - x + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

\frac{u ^{2}}{1 - u ^{2}} = \frac{1}{1 - u ^{2}} - 1

= 2 \cdot \int \frac{1}{1 - u ^{2}} - 1 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

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

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

\int 1 d u = u

= 2 (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u)

Substitute back

u = \frac{x}{2}

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

Simplify

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

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

Add a constant to the solution

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

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