What is the integral of 6xsqrt(x-4) ?

The integral of 6xsqrt(x-4) is 6(2/5 (x-4)^{5/2}+8/3 (x-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 6xsqrt(x-4)

\int 6 x x - 4 d x

6 (\frac{2}{5} (x - 4)^{\frac{5}{2}} + \frac{8}{3} (x - 4)^{\frac{3}{2}}) + C

Solution steps

\int 6 x x - 4 d x

Take the constant out:

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

= 6 \cdot \int x x - 4 d x

Apply u-substitution

: \int (u + 4) u d u

= 6 \cdot \int (u + 4) u d u

Expand

(u + 4) u : u^{\frac{3}{2}} + 4 u

= 6 \cdot \int u^{\frac{3}{2}} + 4 u 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

= 6 (\int u^{\frac{3}{2}} d u + \int 4 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 4 u d u = \frac{8}{3} u^{\frac{3}{2}}

= 6 (\frac{2}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}})

Substitute back

u = x - 4

= 6 (\frac{2}{5} (x - 4)^{\frac{5}{2}} + \frac{8}{3} (x - 4)^{\frac{3}{2}})

Add a constant to the solution

= 6 (\frac{2}{5} (x - 4)^{\frac{5}{2}} + \frac{8}{3} (x - 4)^{\frac{3}{2}}) + C

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

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

\int \frac{z}{e ^{8 z}} d z

\int x (25 - x)^{\frac{7}{2}} d x

\int \frac{cos ( 2 x )}{sin ( x ) - cos ( x )} d x

What is the integral of 6xsqrt(x-4) ?
The integral of 6xsqrt(x-4) is 6(2/5 (x-4)^{5/2}+8/3 (x-4)^{3/2})+C