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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

2 y^{2} + 2 x^{2} + C

Solution steps

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

Take the constant out:

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

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

\frac{x}{2 y ^{2} + 4 x ^{2}} = \frac{1}{2} \frac{x}{y ^{2} + 2 x ^{2}}

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

Take the constant out:

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

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

Apply u-substitution

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

= 4 \cdot \frac{1}{2} \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}{2} \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

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

Substitute back

u = y^{2} + 2 x^{2}

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

Simplify

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

= 2 y^{2} + 2 x^{2}

Add a constant to the solution

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

\int_{0}^{1} 3 e^{x} - 3 x e^{x^{2}} d x

\int \frac{( u )}{( u + 1 ) ( u - 1 )} d u

\frac{d}{d x} (e^{3 x^{3} - x})

x^{4} - x^{3} + 2

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

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