What is the integral of 1/(xsqrt(x^2+36)) ?

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

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integral of 1/(xsqrt(x^2+36))

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

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

Solution steps

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

Apply Trigonometric Substitution

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

= \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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{6} ln ∣ v ∣

Substitute back

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

Add a constant to the solution

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

\int \frac{1}{u ln ( u )} d u

\int 3^{t^{2}} d t

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

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

x \to \infty lim (0.5 3^{1})

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