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

The integral of (y+2x)^{-1/2} is 2(y+2x)^{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 (y+2x)^{-1/2}

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

2 (y + 2 x)^{\frac{1}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{\frac{1}{2}}} d u

= \int \frac{1}{u ^{\frac{1}{2}}} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 2 u^{\frac{1}{2}}

Substitute back

u = y + 2 x

= 2 (y + 2 x)^{\frac{1}{2}}

Add a constant to the solution

= 2 (y + 2 x)^{\frac{1}{2}} + C

\int \frac{5}{1 + 25 x ^{2}} + ln (7) \cdot 7^{x} d x

\int - \frac{2}{2 v - 1} - \frac{1}{v - 1} d v

\int x^{4} + 4 x d x

\int 4 x 1 - 2 x^{2} d x

\int \frac{( cos ( x ) )}{( 4 - sin ^{2} ( x ) )} d x

What is the integral of (y+2x)^{-1/2} ?
The integral of (y+2x)^{-1/2} is 2(y+2x)^{1/2}+C