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

The integral of ((x^3-5x+2))/(x^{1/2)} is 2/7 x^{7/2}-10/3 x^{3/2}+4x^{1/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x^3-5x+2))/(x^{1/2)}

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

\frac{2}{7} x^{\frac{7}{2}} - \frac{10}{3} x^{\frac{3}{2}} + 4 x^{\frac{1}{2}} + C

Solution steps

\int \frac{x ^{3} - 5 x + 2}{x ^{\frac{1}{2}}} d x

Expand

\frac{x ^{3} - 5 x + 2}{x ^{\frac{1}{2}}} : x^{\frac{5}{2}} - 5 x^{\frac{1}{2}} + \frac{2}{x ^{\frac{1}{2}}}

= \int x^{\frac{5}{2}} - 5 x^{\frac{1}{2}} + \frac{2}{x ^{\frac{1}{2}}} 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

= \int x^{\frac{5}{2}} d x - \int 5 x^{\frac{1}{2}} d x + \int \frac{2}{x ^{\frac{1}{2}}} d x

\int x^{\frac{5}{2}} d x = \frac{2}{7} x^{\frac{7}{2}}

\int 5 x^{\frac{1}{2}} d x = \frac{10}{3} x^{\frac{3}{2}}

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

= \frac{2}{7} x^{\frac{7}{2}} - \frac{10}{3} x^{\frac{3}{2}} + 4 x^{\frac{1}{2}}

Add a constant to the solution

= \frac{2}{7} x^{\frac{7}{2}} - \frac{10}{3} x^{\frac{3}{2}} + 4 x^{\frac{1}{2}} + C

\frac{d}{d x} ((4 x^{2} + 3 x)^{2})

\int x^{3} x^{2} + 34 d x

x \to 4 + lim (- \frac{3}{8 - 2 x})

x \to 0 lim (∣ x ∣ sin (x))

y = 5 tan (x), at x = \frac{π}{3}

What is the integral of ((x^3-5x+2))/(x^{1/2)} ?
The integral of ((x^3-5x+2))/(x^{1/2)} is 2/7 x^{7/2}-10/3 x^{3/2}+4x^{1/2}+C