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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(2+4y^2)

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

\frac{1}{2} y 1 + 2 y^{2} + \frac{1}{2} ln 2 y + 1 + 2 y^{2} + C

Solution steps

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

Apply Trigonometric Substitution

: \int sec^{3} (u) d u

= \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{2}{2} y)

= \frac{sec ^{2} ( arctan ( \frac{2}{2} y ) ) sin ( arctan ( \frac{2}{2} y ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{2} y)) + sec (arctan (\frac{2}{2} y))

Simplify

\frac{sec ^{2} ( arctan ( \frac{2}{2} y ) ) sin ( arctan ( \frac{2}{2} y ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{2} y)) + sec (arctan (\frac{2}{2} y)) : \frac{1}{2} y 1 + 2 y^{2} + \frac{1}{2} ln 2 y + 1 + 2 y^{2}

= \frac{1}{2} y 1 + 2 y^{2} + \frac{1}{2} ln 2 y + 1 + 2 y^{2}

Add a constant to the solution

= \frac{1}{2} y 1 + 2 y^{2} + \frac{1}{2} ln 2 y + 1 + 2 y^{2} + C

\frac{d}{d x} (x 49 - x^{2})

\int x^{2} + 5 d x

\int \frac{1}{8 - 6 x - 9 x ^{2}} d x

\frac{x ^{2} + sec ( x )}{x - 2 e ^{x}}

\frac{d}{d x} (\frac{x ^{4}}{8} + \frac{1}{4 y ^{2}})

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