What is the integral of x(x^2-4)^{7/2} ?

The integral of x(x^2-4)^{7/2} is 1/9 (x^2-4)^{9/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x(x^2-4)^{7/2}

\int x (x^{2} - 4)^{\frac{7}{2}} d x

\frac{1}{9} (x^{2} - 4)^{\frac{9}{2}} + C

Solution steps

\int x (x^{2} - 4)^{\frac{7}{2}} d x

Apply u-substitution

: \int \frac{u ^{\frac{7}{2}}}{2} d u

= \int \frac{u ^{\frac{7}{2}}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u^{\frac{7}{2}} d u

Apply the Power Rule

: \frac{2}{9} u^{\frac{9}{2}}

= \frac{1}{2} \cdot \frac{2}{9} u^{\frac{9}{2}}

Substitute back

u = x^{2} - 4

= \frac{1}{2} \cdot \frac{2}{9} (x^{2} - 4)^{\frac{9}{2}}

Simplify

\frac{1}{2} \cdot \frac{2}{9} (x^{2} - 4)^{\frac{9}{2}} : \frac{1}{9} (x^{2} - 4)^{\frac{9}{2}}

= \frac{1}{9} (x^{2} - 4)^{\frac{9}{2}}

Add a constant to the solution

= \frac{1}{9} (x^{2} - 4)^{\frac{9}{2}} + C

\int (1 - 3 x)^{9} d x

\int \frac{x + ln ( 1 - x )}{x ^{2}} d x

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

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

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

What is the integral of x(x^2-4)^{7/2} ?
The integral of x(x^2-4)^{7/2} is 1/9 (x^2-4)^{9/2}+C