What is the integral of 1/(sqrt(9x^2-16)) ?

The integral of 1/(sqrt(9x^2-16)) is 1/3 ln((|sqrt(9x^2-16)+3x|)/4)+C

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integral of 1/(sqrt(9x^2-16))

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

\frac{1}{3} ln (\frac{9 x ^{2} - 16 + 3 x}{4}) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3} d u

= \int \frac{sec ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{3}{4} x)

= \frac{1}{3} ln tan (arcsec (\frac{3}{4} x)) + sec (arcsec (\frac{3}{4} x))

Simplify

\frac{1}{3} ln tan (arcsec (\frac{3}{4} x)) + sec (arcsec (\frac{3}{4} x)) : \frac{1}{3} ln (\frac{9 x ^{2} - 16 + 3 x}{4})

= \frac{1}{3} ln (\frac{9 x ^{2} - 16 + 3 x}{4})

Add a constant to the solution

= \frac{1}{3} ln (\frac{9 x ^{2} - 16 + 3 x}{4}) + C

\frac{\partial}{\partial y} (z arctan (\frac{y}{x}))

\frac{d}{d x} (e^{1 - 3 x})

\int_{1}^{2} \frac{e ^{x}}{x} d x

(7.2) (2.12)

x^{3}, 0, 1

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