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

The integral of x/(sqrt(2+2x^2)) is 1/2 sqrt(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 x/(sqrt(2+2x^2))

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

\frac{1}{2} 2 + 2 x^{2} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

= \frac{1}{2} \cdot \int \frac{1}{2 v} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{v} d v

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot 2 v^{\frac{1}{2}}

Substitute back

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} 2 + 2 x^{2} + C

\frac{1 + x}{4 - x}

\int \frac{3 x + 1}{( x - 4 ) ( x - 2 )} d x

\int e^{c o s (27 t)} sin (27 t) d t

\frac{d}{d x} (\frac{x ^{2} - 8 x + 12}{( x - 4 ) ^{2}})

n = 0 \sum \infty \frac{x ^{n}}{n + 3}

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