What is the integral of (x^2-4)^{1/2} ?

The integral of (x^2-4)^{1/2} is 4(1/8 x(x^2-4)^{1/2}-1/2 ln((|x+(x^2-4)^{1/2}|)/2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2-4)^{1/2}

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

4 \frac{1}{8} x (x^{2} - 4)^{\frac{1}{2}} - \frac{1}{2} ln \frac{x + ( x ^{2} - 4 ) ^{\frac{1}{2}}}{2} + C

Solution steps

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

Apply Trigonometric Substitution

: \int 4 tan^{2} (u) sec (u) d u

= \int 4 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

= 4 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 4 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= 4 \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

= 4 (- \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) ∣

= 4 (- 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}{2} x)

= 4 (- ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x)) + \frac{1}{2} sec (arcsec (\frac{1}{2} x)) tan (arcsec (\frac{1}{2} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x)))

Simplify

4 (- ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x)) + \frac{1}{2} sec (arcsec (\frac{1}{2} x)) tan (arcsec (\frac{1}{2} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x))) : 4 \frac{1}{8} x (x^{2} - 4)^{\frac{1}{2}} - \frac{1}{2} ln \frac{x + ( x ^{2} - 4 ) ^{\frac{1}{2}}}{2}

= 4 \frac{1}{8} x (x^{2} - 4)^{\frac{1}{2}} - \frac{1}{2} ln \frac{x + ( x ^{2} - 4 ) ^{\frac{1}{2}}}{2}

Add a constant to the solution

= 4 \frac{1}{8} x (x^{2} - 4)^{\frac{1}{2}} - \frac{1}{2} ln \frac{x + ( x ^{2} - 4 ) ^{\frac{1}{2}}}{2} + C

\int_{6}^{8} \frac{10}{( x - 6 ) ^{3}} d x

f (x) = x^{2} e^{- 7 x}

\int x^{2} \cdot 2 x d x

\int_{0}^{7} \frac{7 t}{( t - 8 ) ^{2}} d t

\frac{\partial}{\partial x} (sin^{2} (x) cos^{2} (y))

What is the integral of (x^2-4)^{1/2} ?
The integral of (x^2-4)^{1/2} is 4(1/8 x(x^2-4)^{1/2}-1/2 ln((|x+(x^2-4)^{1/2}|)/2))+C