What is the integral of 1/(xsqrt(x^2+9)) ?

The integral of 1/(xsqrt(x^2+9)) is 1/3 ln|tan((arctan(1/3 x))/2)|+C

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integral of 1/(xsqrt(x^2+9))

\int \frac{1}{x x ^{2} + 9} d x

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

Solution steps

\int \frac{1}{x x ^{2} + 9} 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

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\frac{3}{( s + 2 )}

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- t} cos (t)

\frac{d}{d x} (ln (e^{s i n (x)} \cdot x^{c o s (x)}))

What is the integral of 1/(xsqrt(x^2+9)) ?
The integral of 1/(xsqrt(x^2+9)) is 1/3 ln|tan((arctan(1/3 x))/2)|+C