What is the integral of (x^2)/((x^2-8^2)^{3/2)} ?

The integral of (x^2)/((x^2-8^2)^{3/2)} is -x/((x^2-64)^{1/2)}+ln(1/8 |(x^2-64)^{1/2}+x|)+C

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integral of (x^2)/((x^2-8^2)^{3/2)}

\int \frac{x ^{2}}{( x ^{2} - 8 ^{2} ) ^{\frac{3}{2}}} d x

- \frac{x}{( x ^{2} - 64 ) ^{\frac{1}{2}}} + ln (\frac{1}{8} (x^{2} - 64)^{\frac{1}{2}} + x) + C

Solution steps

\int \frac{x ^{2}}{( x ^{2} - 8 ^{2} ) ^{\frac{3}{2}}} d x

Simplify

= \int \frac{x ^{2}}{( x ^{2} - 64 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int sec^{3} (u) cot^{2} (u) d u

= \int sec^{3} (u) cot^{2} (u) d u

Rewrite using trig identities

= \int \frac{csc ^{2} ( u )}{cos ( u )} d u

Apply Integration By Parts

: - csc (u) - \int - sec (u) d u

= - csc (u) - \int - sec (u) d u

\int - sec (u) d u = - ln ∣ tan (u) + sec (u) ∣

= - csc (u) - (- ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{1}{8} x)

= - csc (arcsec (\frac{1}{8} x)) - (- ln tan (arcsec (\frac{1}{8} x)) + sec (arcsec (\frac{1}{8} x)))

Simplify

- csc (arcsec (\frac{1}{8} x)) - (- ln tan (arcsec (\frac{1}{8} x)) + sec (arcsec (\frac{1}{8} x))) : - \frac{x}{( x ^{2} - 64 ) ^{\frac{1}{2}}} + ln (\frac{1}{8} (x^{2} - 64)^{\frac{1}{2}} + x)

= - \frac{x}{( x ^{2} - 64 ) ^{\frac{1}{2}}} + ln (\frac{1}{8} (x^{2} - 64)^{\frac{1}{2}} + x)

Add a constant to the solution

= - \frac{x}{( x ^{2} - 64 ) ^{\frac{1}{2}}} + ln (\frac{1}{8} (x^{2} - 64)^{\frac{1}{2}} + x) + C

\frac{\partial}{\partial x} (1 + x y^{2} - 2 z^{2})

\int \frac{e ^{x}}{e ^{x}} d x

\frac{1}{y ^{2}} - \frac{9}{y ^{4}} (y + 3 y^{3})

n = 1 \sum \infty \frac{10 n}{n + 1}

\frac{d}{d x} (\frac{1}{( - 3 x ^{2} + 3 x - 2 ) ^{3}})

What is the integral of (x^2)/((x^2-8^2)^{3/2)} ?
The integral of (x^2)/((x^2-8^2)^{3/2)} is -x/((x^2-64)^{1/2)}+ln(1/8 |(x^2-64)^{1/2}+x|)+C