What is the integral of x^3sqrt(x^2-25) ?

The integral of x^3sqrt(x^2-25) is 1/5 (x^2-25)^{5/2}+25/3 (x^2-25)^{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^2-25)

\int x^{3} x^{2} - 25 d x

\frac{1}{5} (x^{2} - 25)^{\frac{5}{2}} + \frac{25}{3} (x^{2} - 25)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} - 25 d x

Apply u-substitution

: \int (u^{2} + 25) u^{2} d u

= \int (u^{2} + 25) u^{2} d u

Expand

(u^{2} + 25) u^{2} : u^{4} + 25 u^{2}

= \int u^{4} + 25 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 25 u^{2} d u

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

\int 25 u^{2} d u = \frac{25 u ^{3}}{3}

= \frac{u ^{5}}{5} + \frac{25 u ^{3}}{3}

Substitute back

u = x^{2} - 25

= \frac{( x ^{2} - 25 ) ^{5}}{5} + \frac{25 ( x ^{2} - 25 ) ^{3}}{3}

Simplify

\frac{( x ^{2} - 25 ) ^{5}}{5} + \frac{25 ( x ^{2} - 25 ) ^{3}}{3} : \frac{1}{5} (x^{2} - 25)^{\frac{5}{2}} + \frac{25}{3} (x^{2} - 25)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} - 25)^{\frac{5}{2}} + \frac{25}{3} (x^{2} - 25)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} - 25)^{\frac{5}{2}} + \frac{25}{3} (x^{2} - 25)^{\frac{3}{2}} + C

f (x) = e^{x + 1} - 3 e^{x} + 2 e^{x^{3}}

\int_{2}^{\infty} 2 x^{- 2} d x

\int \frac{1}{3 + 2 y} d y

\int (\frac{1}{y}) d y

\int_{0}^{9} π y^{3} d y

What is the integral of x^3sqrt(x^2-25) ?
The integral of x^3sqrt(x^2-25) is 1/5 (x^2-25)^{5/2}+25/3 (x^2-25)^{3/2}+C