What is the integral of (x^3)/(sqrt(x^2-1)) ?

The integral of (x^3)/(sqrt(x^2-1)) is 1/3 (x^2-1)^{3/2}+sqrt(x^2-1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3)/(sqrt(x^2-1))

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

\frac{1}{3} (x^{2} - 1)^{\frac{3}{2}} + x^{2} - 1 + C

Solution steps

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

Apply u-substitution

: \int u^{2} + 1 d u

= \int u^{2} + 1 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 u^{2} d u + \int 1 d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

= \frac{u ^{3}}{3} + u

Substitute back

u = x^{2} - 1

= \frac{( x ^{2} - 1 ) ^{3}}{3} + x^{2} - 1

Simplify

\frac{( x ^{2} - 1 ) ^{3}}{3} + x^{2} - 1 : \frac{1}{3} (x^{2} - 1)^{\frac{3}{2}} + x^{2} - 1

= \frac{1}{3} (x^{2} - 1)^{\frac{3}{2}} + x^{2} - 1

Add a constant to the solution

= \frac{1}{3} (x^{2} - 1)^{\frac{3}{2}} + x^{2} - 1 + C

25 - x^{2} d x

\frac{\partial}{\partial y} (x e^{2 yz})

\frac{\partial}{\partial s} (\frac{s}{s ^{2} + t ^{2}})

\int x^{2} + 6 x + 3 d x

\frac{d}{d x} ((\frac{x + 1}{x - 1})^{3})

What is the integral of (x^3)/(sqrt(x^2-1)) ?
The integral of (x^3)/(sqrt(x^2-1)) is 1/3 (x^2-1)^{3/2}+sqrt(x^2-1)+C