What is the integral of sqrt(4x^2-256) ?

The integral of sqrt(4x^2-256) is xsqrt(x^2-64)-64ln(1/8 |x+sqrt(x^2-64)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(4x^2-256)

\int 4 x^{2} - 256 d x

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

Solution steps

\int 4 x^{2} - 256 d x

Apply Trigonometric Substitution

: \int 128 tan^{2} (u) sec (u) d u

= \int 128 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

= 128 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 128 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= 128 \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

= 128 (- \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) ∣

= 128 (- 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{2}{16} x)

= 128 (- ln tan (arcsec (\frac{2}{16} x)) + sec (arcsec (\frac{2}{16} x)) + \frac{1}{2} sec (arcsec (\frac{2}{16} x)) tan (arcsec (\frac{2}{16} x)) + \frac{1}{2} ln tan (arcsec (\frac{2}{16} x)) + sec (arcsec (\frac{2}{16} x)))

Simplify

128 (- ln tan (arcsec (\frac{2}{16} x)) + sec (arcsec (\frac{2}{16} x)) + \frac{1}{2} sec (arcsec (\frac{2}{16} x)) tan (arcsec (\frac{2}{16} x)) + \frac{1}{2} ln tan (arcsec (\frac{2}{16} x)) + sec (arcsec (\frac{2}{16} x))) : x x^{2} - 64 - 64 ln (\frac{1}{8} x + x^{2} - 64)

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

Add a constant to the solution

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

\int \frac{x - 1}{x ^{2} - 2 x - 3} d x

\int \frac{1}{y \cdot ln ( y )} d y

\int \frac{130}{( x + 4 ) ^{2}} d x

\int 6 x^{2} (2 x^{3} - 1)^{99} d x

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

What is the integral of sqrt(4x^2-256) ?
The integral of sqrt(4x^2-256) is xsqrt(x^2-64)-64ln(1/8 |x+sqrt(x^2-64)|)+C