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

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

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

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

Factor

x^{2} - 9 : - (- x^{2} + 9)

= \int \frac{x ^{2}}{- ( - x ^{2} + 9 )} d x

Take the constant out:

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

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

Apply Integral Substitution

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

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

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

\int \frac{1}{- u ^{2} + 1} 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

\frac{d}{d x} (- 4^{3})

y = arccos (\frac{9}{x})

f (x) = 4 x - x^{2}

\frac{d}{d x} (50 x)

\int_{25}^{50} 10 d x

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