What is the integral of 1/(xsqrt(4x^2+16)) ?

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

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integral of 1/(xsqrt(4x^2+16))

\int \frac{1}{x 4 x ^{2} + 16} d x

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

Solution steps

\int \frac{1}{x 4 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{2}{4} x )}{2})

Simplify

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

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

Add a constant to the solution

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

x \to 1 lim (e^{7 x^{3} - 7 x})

x \to 5 lim (29 - x^{2})

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

f (x) = (2 x + 3) (x^{2} - 4)

\frac{4 x}{3 - cot ( x )}

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