What is the integral of 2/(xsqrt(36+x^2)) ?

The integral of 2/(xsqrt(36+x^2)) is 1/3 ln|tan(1/2 arctan(1/6 x))|+C

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integral of 2/(xsqrt(36+x^2))

\int \frac{2}{x 36 + x ^{2}} d x

\frac{1}{3} ln tan (\frac{1}{2} arctan (\frac{1}{6} x)) + C

Solution steps

\int \frac{2}{x 36 + x ^{2}} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{x 36 + x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{6 tan ( u )} d u

= 2 \cdot \int \frac{sec ( u )}{6 tan ( u )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

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

= 2 \cdot \frac{1}{6} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{6} ln ∣ v ∣

Substitute back

= 2 \cdot \frac{1}{6} ln tan (\frac{arctan ( \frac{1}{6} x )}{2})

Simplify

2 \cdot \frac{1}{6} ln tan (\frac{arctan ( \frac{1}{6} x )}{2}) : \frac{1}{3} ln tan (\frac{1}{2} arctan (\frac{1}{6} x))

= \frac{1}{3} ln tan (\frac{1}{2} arctan (\frac{1}{6} x))

Add a constant to the solution

= \frac{1}{3} ln tan (\frac{1}{2} arctan (\frac{1}{6} x)) + C

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What is the integral of 2/(xsqrt(36+x^2)) ?
The integral of 2/(xsqrt(36+x^2)) is 1/3 ln|tan(1/2 arctan(1/6 x))|+C