What is the integral of (x^2)/(sqrt(x^2-1)) ?

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

= \frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u

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

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

Substitute back

u = arcsec (x)

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

Simplify

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

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

Add a constant to the solution

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

g (x) = 2 (4 - 7 x)^{4}

\int \frac{sin ( 3 x )}{cos ( 3 x ) + 1} d x

\int_{0}^{1} 2 x e^{x} d x

\int \frac{cos ^{2} ( x )}{1 + sin ( x )} d x

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

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