What is the integral of 1/(xsqrt(9x^2+16)) ?

The integral of 1/(xsqrt(9x^2+16)) is 1/4 ln|tan((arctan(3/4 x))/2)|+C

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integral of 1/(xsqrt(9x^2+16))

\int \frac{1}{x 9 x ^{2} + 16} d x

\frac{1}{4} ln tan (\frac{arctan ( \frac{3}{4} x )}{2}) + C

Solution steps

\int \frac{1}{x 9 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{3}{4} x )}{2})

Add a constant to the solution

= \frac{1}{4} ln tan (\frac{arctan ( \frac{3}{4} x )}{2}) + C

\int t^{2} t^{3} + 1 d t

\int \frac{r ^{3}}{4 + r ^{2}} d r

x \to 2 - lim (\frac{1 - x}{x - 2})

x y^{^{'}} - 2 y = 12 x^{2}

\frac{d}{d x} (x^{2} + 5 x - 8)

What is the integral of 1/(xsqrt(9x^2+16)) ?
The integral of 1/(xsqrt(9x^2+16)) is 1/4 ln|tan((arctan(3/4 x))/2)|+C