What is the integral of 1/(sqrt(2+5y^2)) ?

The integral of 1/(sqrt(2+5y^2)) is 1/(sqrt(5))ln|sqrt(5/2)y+sqrt(1/2 (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 1/(sqrt(2+5y^2))

\int \frac{1}{2 + 5 y ^{2}} d y

\frac{1}{5} ln \frac{5}{2} y + \frac{1}{2} (2 + 5 y^{2}) + C

Solution steps

\int \frac{1}{2 + 5 y ^{2}} d y

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{5} d u

= \int \frac{sec ( u )}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{5} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{5}{2} y)

= \frac{1}{5} ln tan (arctan (\frac{5}{2} y)) + sec (arctan (\frac{5}{2} y))

Simplify

\frac{1}{5} ln tan (arctan (\frac{5}{2} y)) + sec (arctan (\frac{5}{2} y)) : \frac{1}{5} ln \frac{5}{2} y + \frac{1}{2} (2 + 5 y^{2})

= \frac{1}{5} ln \frac{5}{2} y + \frac{1}{2} (2 + 5 y^{2})

Add a constant to the solution

= \frac{1}{5} ln \frac{5}{2} y + \frac{1}{2} (2 + 5 y^{2}) + C

x \to \infty lim (x^{6 x})

5 x - 3

ln (x + 9)

\int (4 x^{3} - 9 x^{2} + 7 x + 3 e) e^{- x} d x

y^{'} = \frac{( x y + y ^{2} )}{x ^{2}}

What is the integral of 1/(sqrt(2+5y^2)) ?
The integral of 1/(sqrt(2+5y^2)) is 1/(sqrt(5))ln|sqrt(5/2)y+sqrt(1/2 (2+5y^2))|+C