What is the integral of 9xsqrt(6x-3) ?

The integral of 9xsqrt(6x-3) is (3sqrt(3))/5 (3x+1)(2x-1)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 9xsqrt(6x-3)

\int 9 x 6 x - 3 d x

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

Solution steps

\int 9 x 6 x - 3 d x

Take the constant out:

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

= 9 \cdot \int x 6 x - 3 d x

Apply u-substitution

= 9 \cdot \int \frac{u ( u + 3 )}{36} d u

Take the constant out:

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

= 9 \cdot \frac{1}{36} \cdot \int u (u + 3) d u

Expand

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

= 9 \cdot \frac{1}{36} \cdot \int u^{\frac{3}{2}} + 3 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 \cdot \frac{1}{36} (\int u^{\frac{3}{2}} d u + \int 3 u d u)

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

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

= 9 \cdot \frac{1}{36} (\frac{2}{5} u^{\frac{5}{2}} + 2 u^{\frac{3}{2}})

Substitute back

u = 6 x - 3

= 9 \cdot \frac{1}{36} (\frac{2}{5} (6 x - 3)^{\frac{5}{2}} + 2 (6 x - 3)^{\frac{3}{2}})

Simplify

9 \cdot \frac{1}{36} (\frac{2}{5} (6 x - 3)^{\frac{5}{2}} + 2 (6 x - 3)^{\frac{3}{2}}) : \frac{3 3}{5} (3 x + 1) (2 x - 1)^{\frac{3}{2}}

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

Add a constant to the solution

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

\int \frac{x ^{2}}{3 + x ^{3}} d x

\int 8 x x^{2} + 1 d x

\int \frac{1}{3 x + 1 + 1} d x

\int cos^{3} (x) cos^{3} (x) d x

\int (5 x^{2} + 10) d x

What is the integral of 9xsqrt(6x-3) ?
The integral of 9xsqrt(6x-3) is (3sqrt(3))/5 (3x+1)(2x-1)^{3/2}+C