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

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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

= 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

= 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

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