What is the integral of (x^2)/(16-x^2) ?

The integral of (x^2)/(16-x^2) is 2ln| x/4+1|-2ln| x/4-1|-x+C

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integral of (x^2)/(16-x^2)

\int \frac{x ^{2}}{16 - x ^{2}} d x

2 ln \frac{x}{4} + 1 - 2 ln \frac{x}{4} - 1 - x + C

Solution steps

\int \frac{x ^{2}}{16 - x ^{2}} d x

Apply Integral Substitution

= \int \frac{4 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

= 4 \cdot \int \frac{u ^{2}}{1 - u ^{2}} d u

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

= 4 \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

= 4 (\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

= 4 (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u)

Substitute back

u = \frac{x}{4}

= 4 (\frac{ln \frac{x}{4} + 1}{2} - \frac{ln \frac{x}{4} - 1}{2} - \frac{x}{4})

Simplify

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

= 2 ln \frac{x}{4} + 1 - 2 ln \frac{x}{4} - 1 - x

Add a constant to the solution

= 2 ln \frac{x}{4} + 1 - 2 ln \frac{x}{4} - 1 - x + C

\int (\frac{( 3 x + 2 )}{19 - 5 x + x ^{2}}) d x

\int \frac{12 x ^{2}}{2 x ^{3} + 1} d x

\frac{8 x + x ^{2}}{( 4 + x ) ^{2}}

\int \frac{x}{4 - 5 x ^{2}} d x

f (x) = \frac{x}{x + 5}

What is the integral of (x^2)/(16-x^2) ?
The integral of (x^2)/(16-x^2) is 2ln| x/4+1|-2ln| x/4-1|-x+C