What is the integral of (x^2)/(9-x^2) ?

The integral of (x^2)/(9-x^2) is 3/2 ln| x/3+1|-3/2 ln| x/3-1|-x+C

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integral of (x^2)/(9-x^2)

\int \frac{x ^{2}}{9 - x ^{2}} d x

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

Solution steps

\int \frac{x ^{2}}{9 - x ^{2}} d x

Apply Integral Substitution

= \int \frac{3 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

= 3 \cdot \int \frac{u ^{2}}{1 - u ^{2}} d u

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

= 3 \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

= 3 (\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

= 3 (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u)

Substitute back

u = \frac{x}{3}

= 3 (\frac{ln \frac{x}{3} + 1}{2} - \frac{ln \frac{x}{3} - 1}{2} - \frac{x}{3})

Simplify

3 (\frac{ln \frac{x}{3} + 1}{2} - \frac{ln \frac{x}{3} - 1}{2} - \frac{x}{3}) : \frac{3}{2} ln \frac{x}{3} + 1 - \frac{3}{2} ln \frac{x}{3} - 1 - x

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

Add a constant to the solution

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

f (t) = 1 + t

y^{^{'}} + 3 x^{2} y = x^{2}

f (x) = x^{4} - 13 x^{2} + 36, at x = - 1

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

\frac{d}{d x} (c^{x})

What is the integral of (x^2)/(9-x^2) ?
The integral of (x^2)/(9-x^2) is 3/2 ln| x/3+1|-3/2 ln| x/3-1|-x+C