What is the integral of 1/(xsqrt(3x^2+5)) ?

The integral of 1/(xsqrt(3x^2+5)) is 1/(sqrt(5))ln|tan(1/2 arctan(sqrt(3/5)x))|+C

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integral of 1/(xsqrt(3x^2+5))

\int \frac{1}{x 3 x ^{2} + 5} d x

\frac{1}{5} ln tan (\frac{1}{2} arctan (\frac{3}{5} x)) + C

Solution steps

\int \frac{1}{x 3 x ^{2} + 5} d x

Apply Trigonometric Substitution

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

Apply u-substitution

= \frac{1}{5} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{5} ln ∣ v ∣

Substitute back

= \frac{1}{5} ln tan \frac{arctan ( \frac{3}{5} x )}{2}

Simplify

\frac{1}{5} ln tan \frac{arctan ( \frac{3}{5} x )}{2} : \frac{1}{5} ln tan (\frac{1}{2} arctan (\frac{3}{5} x))

= \frac{1}{5} ln tan (\frac{1}{2} arctan (\frac{3}{5} x))

Add a constant to the solution

= \frac{1}{5} ln tan (\frac{1}{2} arctan (\frac{3}{5} x)) + C

What is the integral of 1/(xsqrt(3x^2+5)) ?
The integral of 1/(xsqrt(3x^2+5)) is 1/(sqrt(5))ln|tan(1/2 arctan(sqrt(3/5)x))|+C