What is the integral of 9xsqrt(x-7) ?

The integral of 9xsqrt(x-7) is 9(2/5 (x-7)^{5/2}+14/3 (x-7)^{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 9xsqrt(x-7)

\int 9 x x - 7 d x

9 (\frac{2}{5} (x - 7)^{\frac{5}{2}} + \frac{14}{3} (x - 7)^{\frac{3}{2}}) + C

Solution steps

\int 9 x x - 7 d x

Take the constant out:

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

= 9 \cdot \int x x - 7 d x

Apply u-substitution

: \int (u + 7) u d u

= 9 \cdot \int (u + 7) u d u

Expand

(u + 7) u : u^{\frac{3}{2}} + 7 u

= 9 \cdot \int u^{\frac{3}{2}} + 7 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

= 9 (\int u^{\frac{3}{2}} d u + \int 7 u d u)

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

\int 7 u d u = \frac{14}{3} u^{\frac{3}{2}}

= 9 (\frac{2}{5} u^{\frac{5}{2}} + \frac{14}{3} u^{\frac{3}{2}})

Substitute back

u = x - 7

= 9 (\frac{2}{5} (x - 7)^{\frac{5}{2}} + \frac{14}{3} (x - 7)^{\frac{3}{2}})

Add a constant to the solution

= 9 (\frac{2}{5} (x - 7)^{\frac{5}{2}} + \frac{14}{3} (x - 7)^{\frac{3}{2}}) + C

\int 8 \cdot 2^{- x} - 4 d x

\int \frac{in ( x )}{x} d x

\int (5 x - x) d x

\int (4 x^{3} + 6 x^{2} + 3) d x

\int \frac{3}{3 + x} d x

What is the integral of 9xsqrt(x-7) ?
The integral of 9xsqrt(x-7) is 9(2/5 (x-7)^{5/2}+14/3 (x-7)^{3/2})+C