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

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

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

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

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

Solution steps

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

\int 4 x e^{3 x} d x

e^{x} + e^{- x}

\frac{\partial}{\partial x} (cos (x + e^{y}))

\int \frac{9 x}{x - 1} d x

y = 3 x^{2} - 12 x + 1, (0, 1)

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