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

The integral of x^3sqrt(x^2-4) is 1/5 (x^2-4)^{5/2}+4/3 (x^2-4)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^3sqrt(x^2-4)

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

\frac{1}{5} (x^{2} - 4)^{\frac{5}{2}} + \frac{4}{3} (x^{2} - 4)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

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

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

Expand

(u^{2} + 4) u^{2} : u^{4} + 4 u^{2}

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

\int u^{4} d u = \frac{u ^{5}}{5}

\int 4 u^{2} d u = \frac{4 u ^{3}}{3}

= \frac{u ^{5}}{5} + \frac{4 u ^{3}}{3}

Substitute back

u = x^{2} - 4

= \frac{( x ^{2} - 4 ) ^{5}}{5} + \frac{4 ( x ^{2} - 4 ) ^{3}}{3}

Simplify

\frac{( x ^{2} - 4 ) ^{5}}{5} + \frac{4 ( x ^{2} - 4 ) ^{3}}{3} : \frac{1}{5} (x^{2} - 4)^{\frac{5}{2}} + \frac{4}{3} (x^{2} - 4)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} - 4)^{\frac{5}{2}} + \frac{4}{3} (x^{2} - 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} - 4)^{\frac{5}{2}} + \frac{4}{3} (x^{2} - 4)^{\frac{3}{2}} + C

7 x^{6} - x^{e}

- 1

\int \frac{1}{x ^{2} + 9} d x

cos (2 x) - 2 x sin (2 x)

\int - \frac{1}{4} e^{- 4 x} d x

What is the integral of x^3sqrt(x^2-4) ?
The integral of x^3sqrt(x^2-4) is 1/5 (x^2-4)^{5/2}+4/3 (x^2-4)^{3/2}+C