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

The integral of xsqrt(1+1/4 x) is 1/15 (3x-8)(x+4)^{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(1+1/4 x)

\int x 1 + \frac{1}{4} x d x

\frac{1}{15} (3 x - 8) (x + 4)^{\frac{3}{2}} + C

Solution steps

\int x 1 + \frac{1}{4} x d x

Simplify

x 1 + \frac{1}{4} x : \frac{1}{2} x 4 + x

= \int \frac{1}{2} x 4 + x d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x 4 + x d x

Apply u-substitution

: \int (u - 4) u d u

= \frac{1}{2} \cdot \int (u - 4) u d u

Expand

(u - 4) u : u^{\frac{3}{2}} - 4 u

= \frac{1}{2} \cdot \int u^{\frac{3}{2}} - 4 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}{2} (\int u^{\frac{3}{2}} d u - \int 4 u d u)

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

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

= \frac{1}{2} (\frac{2}{5} u^{\frac{5}{2}} - \frac{8}{3} u^{\frac{3}{2}})

Substitute back

u = 4 + x

= \frac{1}{2} (\frac{2}{5} (4 + x)^{\frac{5}{2}} - \frac{8}{3} (4 + x)^{\frac{3}{2}})

Simplify

\frac{1}{2} (\frac{2}{5} (4 + x)^{\frac{5}{2}} - \frac{8}{3} (4 + x)^{\frac{3}{2}}) : \frac{1}{15} (3 x - 8) (x + 4)^{\frac{3}{2}}

= \frac{1}{15} (3 x - 8) (x + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{15} (3 x - 8) (x + 4)^{\frac{3}{2}} + C

\int t + sin (t) d t

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

\int \frac{e ^{\frac{x}{2}}}{x} d x

\int 4 x^{2} + 4 x - 3 d x

\int (x^{b} - x^{- b})^{2} d x

What is the integral of xsqrt(1+1/4 x) ?
The integral of xsqrt(1+1/4 x) is 1/15 (3x-8)(x+4)^{3/2}+C