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

The integral of x^2sqrt(x^2-9) is

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

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

81 (\frac{1}{243} x^{3} x^{2} - 9 - \frac{1}{972} x^{3} x^{2} - 9 - \frac{1}{72} x x^{2} - 9 - \frac{1}{8} ln (\frac{1}{3} x^{2} - 9 + x)) + C

Solution steps

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

Multiply by the conjugate of

x^{2} - 9 : \frac{x ^{4} - 81}{x ^{2} + 9}

= \int x^{2} \frac{x ^{4} - 81}{x ^{2} + 9} d x

Apply Integral Substitution

= \int 81 u^{2} (u + 1) (u - 1) d u

Take the constant out:

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

= 81 \cdot \int u^{2} (u + 1) (u - 1) d u

Multiply by the conjugate of

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

= 81 \cdot \int u^{2} \frac{u ^{2} - 1}{u - 1} (u - 1) d u

Multiply

\frac{u ^{2} - 1}{u - 1} (u - 1) : u^{2} - 1

= 81 \cdot \int u^{2} u^{2} - 1 d u

Apply Integration By Parts

= 81 (\frac{1}{3} u^{3} u^{2} - 1 - \int \frac{u ^{4}}{3 u ^{2} - 1} d u)

\int \frac{u ^{4}}{3 u ^{2} - 1} d u = \frac{1}{12} u^{3} u^{2} - 1 + \frac{1}{8} (u u^{2} - 1 + ln u^{2} - 1 + u)

= 81 (\frac{1}{3} u^{3} u^{2} - 1 - (\frac{1}{12} u^{3} u^{2} - 1 + \frac{1}{8} (u u^{2} - 1 + ln u^{2} - 1 + u)))

Substitute back

u = \frac{x}{3}

= 81 (\frac{1}{3} (\frac{x}{3})^{3} (\frac{x}{3})^{2} - 1 - (\frac{1}{12} (\frac{x}{3})^{3} (\frac{x}{3})^{2} - 1 + \frac{1}{8} (\frac{x}{3} (\frac{x}{3})^{2} - 1 + ln (\frac{x}{3})^{2} - 1 + \frac{x}{3})))

Simplify

81 (\frac{1}{3} (\frac{x}{3})^{3} (\frac{x}{3})^{2} - 1 - (\frac{1}{12} (\frac{x}{3})^{3} (\frac{x}{3})^{2} - 1 + \frac{1}{8} (\frac{x}{3} (\frac{x}{3})^{2} - 1 + ln (\frac{x}{3})^{2} - 1 + \frac{x}{3}))) : 81 (\frac{1}{243} x^{3} x^{2} - 9 - \frac{1}{972} x^{3} x^{2} - 9 - \frac{1}{72} x x^{2} - 9 - \frac{1}{8} ln (\frac{1}{3} x^{2} - 9 + x))

= 81 (\frac{1}{243} x^{3} x^{2} - 9 - \frac{1}{972} x^{3} x^{2} - 9 - \frac{1}{72} x x^{2} - 9 - \frac{1}{8} ln (\frac{1}{3} x^{2} - 9 + x))

Add a constant to the solution

= 81 (\frac{1}{243} x^{3} x^{2} - 9 - \frac{1}{972} x^{3} x^{2} - 9 - \frac{1}{72} x x^{2} - 9 - \frac{1}{8} ln (\frac{1}{3} x^{2} - 9 + x)) + C