What is the integral of (4y)/(sqrt(2y^2+1)) ?

The integral of (4y)/(sqrt(2y^2+1)) is 2sqrt(2y^2+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4y)/(sqrt(2y^2+1))

\int \frac{4 y}{2 y ^{2} + 1} d y

2 2 y^{2} + 1 + C

Solution steps

\int \frac{4 y}{2 y ^{2} + 1} d y

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{4 u} d u

= 4 \cdot \int \frac{1}{4 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= 4 \cdot \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 2 y^{2} + 1

= 4 \cdot \frac{1}{4} \cdot 2 (2 y^{2} + 1)^{\frac{1}{2}}

Simplify

4 \cdot \frac{1}{4} \cdot 2 (2 y^{2} + 1)^{\frac{1}{2}} : 2 2 y^{2} + 1

= 2 2 y^{2} + 1

Add a constant to the solution

= 2 2 y^{2} + 1 + C

\int \frac{x - 1}{5 - 4 x - x ^{2}} d x

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

\int x^{3} - x + 1 d x

\int θ^{2} sec^{2} (θ^{3}) d θ

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

What is the integral of (4y)/(sqrt(2y^2+1)) ?
The integral of (4y)/(sqrt(2y^2+1)) is 2sqrt(2y^2+1)+C