What is the integral of xsqrt(4x-3) ?

The integral of xsqrt(4x-3) is 1/20 (2x+1)(4x-3)^{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 xsqrt(4x-3)

\int x 4 x - 3 d x

\frac{1}{20} (2 x + 1) (4 x - 3)^{\frac{3}{2}} + C

Solution steps

\int x 4 x - 3 d x

Apply u-substitution

: \int \frac{u ( u + 3 )}{16} d u

= \int \frac{u ( u + 3 )}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int u (u + 3) d u

Expand

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

= \frac{1}{16} \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

= \frac{1}{16} (\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}}

= \frac{1}{16} (\frac{2}{5} u^{\frac{5}{2}} + 2 u^{\frac{3}{2}})

Substitute back

u = 4 x - 3

= \frac{1}{16} (\frac{2}{5} (4 x - 3)^{\frac{5}{2}} + 2 (4 x - 3)^{\frac{3}{2}})

Simplify

\frac{1}{16} (\frac{2}{5} (4 x - 3)^{\frac{5}{2}} + 2 (4 x - 3)^{\frac{3}{2}}) : \frac{1}{20} (2 x + 1) (4 x - 3)^{\frac{3}{2}}

= \frac{1}{20} (2 x + 1) (4 x - 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{20} (2 x + 1) (4 x - 3)^{\frac{3}{2}} + C

\int \frac{3}{u} d u

\int \frac{- 8}{x} d x

\int 2 x \cdot 1 - x^{2} d x

\int sin (6 t) sec^{2} (cos (6 t)) d t

What is the integral of xsqrt(4x-3) ?
The integral of xsqrt(4x-3) is 1/20 (2x+1)(4x-3)^{3/2}+C