What is the integral of 4xsqrt(x-1) ?

The integral of 4xsqrt(x-1) is 4(2/5 (x-1)^{5/2}+2/3 (x-1)^{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 4xsqrt(x-1)

\int 4 x x - 1 d x

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

Solution steps

\int 4 x x - 1 d x

Take the constant out:

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

= 4 \cdot \int x x - 1 d x

Apply u-substitution

: \int (u + 1) u d u

= 4 \cdot \int (u + 1) u d u

Expand

(u + 1) u : u^{\frac{3}{2}} + u

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

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

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

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

= 4 (\frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = x - 1

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

Add a constant to the solution

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

\int \frac{1}{4 x ^{2} + 4 x + 17} d x

\int sin^{4} (x) \cdot cos^{3} (x) d x

\int \frac{ln ( 1 - x )}{e ^{2} 6 1 - x} d x

\int \frac{x ^{4} + 2 x + 1}{( x ^{2} + 1 ) ( x ^{2} + 9 )} d x

\int x (e^{- x}) d x

What is the integral of 4xsqrt(x-1) ?
The integral of 4xsqrt(x-1) is 4(2/5 (x-1)^{5/2}+2/3 (x-1)^{3/2})+C