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

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

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

\frac{1}{5} ln 5 x + 1 + 5 x^{2} + C

Solution steps

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

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 (5 x)

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

Simplify

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

= \frac{1}{5} ln 5 x + 1 + 5 x^{2}

Add a constant to the solution

= \frac{1}{5} ln 5 x + 1 + 5 x^{2} + C

\int 4 sin^{3} (x co) s^{2} x d x

y = x + 225, (0, 15)

\int 2 x^{2} - 2 d x

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

\frac{d}{d x} (x^{2} - x^{3} + 3)

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