What is the integral of 5/(sqrt(9x^2-16)) ?

The integral of 5/(sqrt(9x^2-16)) is 5/3 ln((|sqrt(9x^2-16)+3x|)/4)+C

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integral of 5/(sqrt(9x^2-16))

\int \frac{5}{9 x ^{2} - 16} d x

\frac{5}{3} ln (\frac{9 x ^{2} - 16 + 3 x}{4}) + C

Solution steps

\int \frac{5}{9 x ^{2} - 16} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{9 x ^{2} - 16} d x

Apply Trigonometric Substitution

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

= 5 \cdot \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

= 5 \cdot \frac{1}{3} \cdot \int sec (u) d u

Use the common integral:

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

= 5 \cdot \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= 5 \cdot \frac{1}{3} ln tan (arcsec (\frac{3}{4} x)) + sec (arcsec (\frac{3}{4} x))

Simplify

5 \cdot \frac{1}{3} ln tan (arcsec (\frac{3}{4} x)) + sec (arcsec (\frac{3}{4} x)) : \frac{5}{3} ln (\frac{9 x ^{2} - 16 + 3 x}{4})

= \frac{5}{3} ln (\frac{9 x ^{2} - 16 + 3 x}{4})

Add a constant to the solution

= \frac{5}{3} ln (\frac{9 x ^{2} - 16 + 3 x}{4}) + C

f (x) = 6 x^{3} + 7

\int 6 e^{6 x} sin (e^{6 x}) d x

\int \frac{1}{4 y ^{2}} d y

\frac{x + 1}{x + 3}

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

What is the integral of 5/(sqrt(9x^2-16)) ?
The integral of 5/(sqrt(9x^2-16)) is 5/3 ln((|sqrt(9x^2-16)+3x|)/4)+C