What is the integral of 1/(xsqrt(x^2+4)) ?

The integral of 1/(xsqrt(x^2+4)) is 1/2 ln|tan((arctan(1/2 x))/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/(xsqrt(x^2+4))

\int \frac{1}{x x ^{2} + 4} d x

\frac{1}{2} ln tan (\frac{arctan ( \frac{1}{2} x )}{2}) + C

Solution steps

\int \frac{1}{x x ^{2} + 4} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{2 tan ( u )} d u

= \int \frac{sec ( u )}{2 tan ( 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{sec ( u )}{tan ( u )} d u

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

= \frac{1}{2} ln tan (\frac{arctan ( \frac{1}{2} x )}{2})

Add a constant to the solution

= \frac{1}{2} ln tan (\frac{arctan ( \frac{1}{2} x )}{2}) + C

f (x) = \frac{x}{9 - x}

\int 9 x 1 - x^{4} d x

\frac{\partial}{\partial x} (ln (x^{2} + y^{2} + 8))

\frac{d}{d x} (\frac{3}{x ^{6}})

2 cos (x) d y = (y sin (x) - y^{3}) d x

What is the integral of 1/(xsqrt(x^2+4)) ?
The integral of 1/(xsqrt(x^2+4)) is 1/2 ln|tan((arctan(1/2 x))/2)|+C