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

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

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

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

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

Solution steps

\int \frac{1}{x 9 + x ^{2}} 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{\partial}{\partial x} (x y - ln (x^{2} - y^{2}))

5 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

\frac{\partial}{\partial x} ((x^{2} + y^{2} + z^{2})^{- \frac{1}{4}})

x \to 0 lim (\frac{x - 2}{x})

\frac{\partial}{\partial y} (- 8 y^{3} + 4 x y + 4 x^{2} + 9 y^{2})

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