What is the integral of (2x)/(sqrt(2x^2+5y^2)) ?

The integral of (2x)/(sqrt(2x^2+5y^2)) is sqrt(2x^2+5y^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x)/(sqrt(2x^2+5y^2))

\int \frac{2 x}{2 x ^{2} + 5 y ^{2}} d x

2 x^{2} + 5 y^{2} + C

Solution steps

\int \frac{2 x}{2 x ^{2} + 5 y ^{2}} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{2 x ^{2} + 5 y ^{2}} d x

Apply u-substitution

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

= 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

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

Apply the Power Rule

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

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

Substitute back

u = 2 x^{2} + 5 y^{2}

= 2 \cdot \frac{1}{4} \cdot 2 (2 x^{2} + 5 y^{2})^{\frac{1}{2}}

Simplify

2 \cdot \frac{1}{4} \cdot 2 (2 x^{2} + 5 y^{2})^{\frac{1}{2}} : 2 x^{2} + 5 y^{2}

= 2 x^{2} + 5 y^{2}

Add a constant to the solution

= 2 x^{2} + 5 y^{2} + C

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

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

\int \frac{1}{x ^{1.3}} d x

\int_{- 3}^{0} (1 + 9 - x^{2}) d x

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x ^{4}})

What is the integral of (2x)/(sqrt(2x^2+5y^2)) ?
The integral of (2x)/(sqrt(2x^2+5y^2)) is sqrt(2x^2+5y^2)+C