What is the integral of 1/(sqrt(9x^2-49)) ?

The integral of 1/(sqrt(9x^2-49)) is 1/3 ln((|sqrt(9x^2-49)+3x|)/7)+C

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integral of 1/(sqrt(9x^2-49))

\int \frac{1}{9 x ^{2} - 49} d x

\frac{1}{3} ln (\frac{9 x ^{2} - 49 + 3 x}{7}) + C

Solution steps

\int \frac{1}{9 x ^{2} - 49} 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}{7} x)

= \frac{1}{3} ln tan (arcsec (\frac{3}{7} x)) + sec (arcsec (\frac{3}{7} x))

Simplify

\frac{1}{3} ln tan (arcsec (\frac{3}{7} x)) + sec (arcsec (\frac{3}{7} x)) : \frac{1}{3} ln (\frac{9 x ^{2} - 49 + 3 x}{7})

= \frac{1}{3} ln (\frac{9 x ^{2} - 49 + 3 x}{7})

Add a constant to the solution

= \frac{1}{3} ln (\frac{9 x ^{2} - 49 + 3 x}{7}) + C

\frac{d}{d x} ((x^{2} - 1)^{3})

\frac{1}{t ^{2} - 1}

\int \frac{7 x - 1}{x ^{2} - 1} d x

f (x) = 4 x + 1

\int \frac{( x ^{3} + 16 )}{x ^{2} + 16} d x

What is the integral of 1/(sqrt(9x^2-49)) ?
The integral of 1/(sqrt(9x^2-49)) is 1/3 ln((|sqrt(9x^2-49)+3x|)/7)+C