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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(x^2-4))/(x^2)

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Simplify

tan^{2} (u) cos (u) : sin (u) tan (u)

= \int sin (u) tan (u) d u

Rewrite using trig identities

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

Expand

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

= \int \frac{1}{cos ( u )} - cos (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

= \int \frac{1}{cos ( u )} d u - \int cos (u) d u

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

\int cos (u) d u = sin (u)

= ln ∣ tan (u) + sec (u) ∣ - sin (u)

Substitute back

u = arcsec (\frac{1}{2} x)

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

Simplify

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

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

Add a constant to the solution

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

\int (\frac{3}{x} + x) d x

y = 2 x, y = 2 x^{3}

\int_{0}^{\frac{13}{12}} \frac{68}{3} x d x

y = 2 x^{2} - 9 x + 10

\frac{d}{d x} (17.5 (x + 1)^{0.75})

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