What is the integral of sqrt(x^2-64) ?

The integral of sqrt(x^2-64) is 64(1/128 xsqrt(x^2-64)-1/2 ln((|x+sqrt(x^2-64)|)/8))+C

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integral of sqrt(x^2-64)

\int x^{2} - 64 d x

64 (\frac{1}{128} x x^{2} - 64 - \frac{1}{2} ln (\frac{x + x ^{2} - 64}{8})) + C

Solution steps

\int x^{2} - 64 d x

Apply Trigonometric Substitution

: \int 64 tan^{2} (u) sec (u) d u

= \int 64 tan^{2} (u) sec (u) d u

Take the constant out:

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

= 64 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 64 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 64 \cdot \int - sec (u) + sec^{3} (u) 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

= 64 (- \int sec (u) d u + \int sec^{3} (u) d u)

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

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= 64 (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

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

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

Simplify

64 (- ln tan (arcsec (\frac{1}{8} x)) + sec (arcsec (\frac{1}{8} x)) + \frac{1}{2} sec (arcsec (\frac{1}{8} x)) tan (arcsec (\frac{1}{8} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{8} x)) + sec (arcsec (\frac{1}{8} x))) : 64 (\frac{1}{128} x x^{2} - 64 - \frac{1}{2} ln (\frac{x + x ^{2} - 64}{8}))

= 64 (\frac{1}{128} x x^{2} - 64 - \frac{1}{2} ln (\frac{x + x ^{2} - 64}{8}))

Add a constant to the solution

= 64 (\frac{1}{128} x x^{2} - 64 - \frac{1}{2} ln (\frac{x + x ^{2} - 64}{8})) + C

\frac{d}{d a} (\frac{π ( 9 a ^{2} + 6 a + 1 )}{54 a})

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

x \to π lim ((π - x) csc (x))

9 x y^{2} - 3 + (9 x^{2} y + 4) \frac{d y}{d x} = 0

\frac{d y}{d x} + 4 y = 6

What is the integral of sqrt(x^2-64) ?
The integral of sqrt(x^2-64) is 64(1/128 xsqrt(x^2-64)-1/2 ln((|x+sqrt(x^2-64)|)/8))+C