What is the integral of (x^3)/(sqrt(x^2-4)) ?

The integral of (x^3)/(sqrt(x^2-4)) is 1/3 (x^2-4)^{3/2}+4sqrt(x^2-4)+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^2-4))

\int \frac{x ^{3}}{x ^{2} - 4} d x

\frac{1}{3} (x^{2} - 4)^{\frac{3}{2}} + 4 x^{2} - 4 + C

Solution steps

\int \frac{x ^{3}}{x ^{2} - 4} d x

Apply u-substitution

: \int u^{2} + 4 d u

= \int u^{2} + 4 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 4 d u

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

\int 4 d u = 4 u

= \frac{u ^{3}}{3} + 4 u

Substitute back

u = x^{2} - 4

= \frac{( x ^{2} - 4 ) ^{3}}{3} + 4 x^{2} - 4

Simplify

\frac{( x ^{2} - 4 ) ^{3}}{3} + 4 x^{2} - 4 : \frac{1}{3} (x^{2} - 4)^{\frac{3}{2}} + 4 x^{2} - 4

= \frac{1}{3} (x^{2} - 4)^{\frac{3}{2}} + 4 x^{2} - 4

Add a constant to the solution

= \frac{1}{3} (x^{2} - 4)^{\frac{3}{2}} + 4 x^{2} - 4 + C

x \to 1 + lim (\frac{( 2 )}{x - 1})

\int_{- 2}^{3} (x^{2} - 4) d x

\int \frac{2 x}{x ^{2} + 6 x + 13} d x

n = 1 \sum \infty \frac{1}{2} n

\int sin (7 x) cos (5 x) d x

What is the integral of (x^3)/(sqrt(x^2-4)) ?
The integral of (x^3)/(sqrt(x^2-4)) is 1/3 (x^2-4)^{3/2}+4sqrt(x^2-4)+C