What is the integral of 1/(x(sqrt(16+x^2))) ?

The integral of 1/(x(sqrt(16+x^2))) is 1/4 ln|tan((arctan(1/4 x))/2)|+C

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integral of 1/(x(sqrt(16+x^2)))

\int \frac{1}{x ( 16 + x ^{2} )} d x

\frac{1}{4} ln tan (\frac{arctan ( \frac{1}{4} x )}{2}) + C

Solution steps

\int \frac{1}{x 16 + x ^{2}} d x

Apply Trigonometric Substitution

= \int \frac{sec ( u )}{4 tan ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

= \frac{1}{4} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{4} ln ∣ v ∣

Substitute back

= \frac{1}{4} ln tan (\frac{arctan ( \frac{1}{4} x )}{2})

Add a constant to the solution

= \frac{1}{4} ln tan (\frac{arctan ( \frac{1}{4} x )}{2}) + C

What is the integral of 1/(x(sqrt(16+x^2))) ?
The integral of 1/(x(sqrt(16+x^2))) is 1/4 ln|tan((arctan(1/4 x))/2)|+C