What is the integral of 1/(xsqrt(x^2+3)) ?

The integral of 1/(xsqrt(x^2+3)) is 1/(sqrt(3))ln|tan((arctan(1/(sqrt(3))x))/2)|+C

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integral of 1/(xsqrt(x^2+3))

\int \frac{1}{x x ^{2} + 3} d x

\frac{1}{3} ln tan \frac{arctan ( \frac{1}{3} x )}{2} + C

Solution steps

\int \frac{1}{x x ^{2} + 3} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3 tan ( u )} d u

= \int \frac{sec ( u )}{3 tan ( u )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

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

= \frac{1}{3} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} ln ∣ v ∣

Substitute back

= \frac{1}{3} ln tan \frac{arctan ( \frac{1}{3} x )}{2}

Add a constant to the solution

= \frac{1}{3} ln tan \frac{arctan ( \frac{1}{3} x )}{2} + C

\frac{d}{d x} (\frac{x cos ( x ) + π}{x - π})

x \to 4 lim (7 x + 12 x)

x \to 0 lim ((1 - e^{4 x})^{x})

n = 1 \sum \infty n^{- \frac{1}{6}}

\frac{d}{d x} (sin^{3} (cos (2 x)))

What is the integral of 1/(xsqrt(x^2+3)) ?
The integral of 1/(xsqrt(x^2+3)) is 1/(sqrt(3))ln|tan((arctan(1/(sqrt(3))x))/2)|+C