What is the integral of ((5y+4))/(sqrt(5y-1)) ?

The integral of ((5y+4))/(sqrt(5y-1)) is 2/15 (5y-1)^{3/2}+2sqrt(5y-1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((5y+4))/(sqrt(5y-1))

\int \frac{( 5 y + 4 )}{5 y - 1} d y

\frac{2}{15} (5 y - 1)^{\frac{3}{2}} + 2 5 y - 1 + C

Solution steps

\int \frac{5 y + 4}{5 y - 1} d y

Apply u-substitution

: \int \frac{2 ( u ^{2} + 5 )}{5} d u

= \int \frac{2 ( u ^{2} + 5 )}{5} d u

Take the constant out:

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

= \frac{2}{5} \cdot \int u^{2} + 5 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{2}{5} (\int u^{2} d u + \int 5 d u)

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

\int 5 d u = 5 u

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

Substitute back

u = 5 y - 1

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

Simplify

\frac{2}{5} (\frac{( 5 y - 1 ) ^{3}}{3} + 5 5 y - 1) : \frac{2}{15} (5 y - 1)^{\frac{3}{2}} + 2 5 y - 1

= \frac{2}{15} (5 y - 1)^{\frac{3}{2}} + 2 5 y - 1

Add a constant to the solution

= \frac{2}{15} (5 y - 1)^{\frac{3}{2}} + 2 5 y - 1 + C

\int \frac{sin ^{2} ( x )}{e ^{2 x}} d x

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

\int \frac{2 tan ( x )}{2 + 3 cos ( x )} d x

\int \frac{7}{2 x 9 x ^{2} - 1} d x

\int x^{5} (8 + x^{3})^{\frac{3}{2}} d x

What is the integral of ((5y+4))/(sqrt(5y-1)) ?
The integral of ((5y+4))/(sqrt(5y-1)) is 2/15 (5y-1)^{3/2}+2sqrt(5y-1)+C