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

The integral of ((2x-3)^2)/(3sqrt(x)) is 1/3 (8/5 x^{5/2}-8x^{3/2}+18sqrt(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((2x-3)^2)/(3sqrt(x))

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

\frac{1}{3} (\frac{8}{5} x^{\frac{5}{2}} - 8 x^{\frac{3}{2}} + 18 x) + C

Solution steps

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

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{( 2 x - 3 ) ^{2}}{x} d x

Expand

\frac{( 2 x - 3 ) ^{2}}{x} : 4 x^{\frac{3}{2}} - 12 x + \frac{9}{x}

= \frac{1}{3} \cdot \int 4 x^{\frac{3}{2}} - 12 x + \frac{9}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int 4 x^{\frac{3}{2}} d x - \int 12 x d x + \int \frac{9}{x} d x)

\int 4 x^{\frac{3}{2}} d x = \frac{8}{5} x^{\frac{5}{2}}

\int 12 x d x = 8 x^{\frac{3}{2}}

\int \frac{9}{x} d x = 18 x

= \frac{1}{3} (\frac{8}{5} x^{\frac{5}{2}} - 8 x^{\frac{3}{2}} + 18 x)

Add a constant to the solution

= \frac{1}{3} (\frac{8}{5} x^{\frac{5}{2}} - 8 x^{\frac{3}{2}} + 18 x) + C

\int \frac{1}{2 x ^{2} - 8 x + 10} d x

\int (x^{1.4} + 11 x^{4.5}) d x

\int e^{- s x} sin (2 x) d x

\int (x^{2} + x + 1) \cdot e^{3 x} d x

\int - 0.6 x + 55 d x

What is the integral of ((2x-3)^2)/(3sqrt(x)) ?
The integral of ((2x-3)^2)/(3sqrt(x)) is 1/3 (8/5 x^{5/2}-8x^{3/2}+18sqrt(x))+C