What is the integral of (2y^2-3y+5)/(sqrt(y)) ?

The integral of (2y^2-3y+5)/(sqrt(y)) is 4/5 y^{5/2}-2y^{3/2}+10sqrt(y)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2y^2-3y+5)/(sqrt(y))

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

\frac{4}{5} y^{\frac{5}{2}} - 2 y^{\frac{3}{2}} + 10 y + C

Solution steps

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

Expand

\frac{2 y ^{2} - 3 y + 5}{y} : 2 y^{\frac{3}{2}} - 3 y + \frac{5}{y}

= \int 2 y^{\frac{3}{2}} - 3 y + \frac{5}{y} d y

Apply the Sum Rule:

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

= \int 2 y^{\frac{3}{2}} d y - \int 3 y d y + \int \frac{5}{y} d y

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

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

\int \frac{5}{y} d y = 10 y

= \frac{4}{5} y^{\frac{5}{2}} - 2 y^{\frac{3}{2}} + 10 y

Add a constant to the solution

= \frac{4}{5} y^{\frac{5}{2}} - 2 y^{\frac{3}{2}} + 10 y + C

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

\int \frac{1}{( y - y ^{2} )} d y

\int (3^{x} - 5^{x}) d x

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

\int cos^{3} (3 x) sin^{\frac{1}{3}} (3 x) d x

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