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

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

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integral of 1/(xsqrt(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

= \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

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

= \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

\frac{d y}{d x} = (x + y - 1)^{2}

\int x^{3} sin (n x) d x

x \frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} = 0

\int_{3}^{5} \frac{x}{x ^{2} + 6 x + 13} d x

x \to 0 lim (\frac{- 2 x ( x + 2 )}{4 x ^{2} - 3 x - 7})

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