What is the integral of 1/(sqrt(36x^2-49)) ?

The integral of 1/(sqrt(36x^2-49)) is 1/6 ln((|sqrt(36x^2-49)+6x|)/7)+C

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integral of 1/(sqrt(36x^2-49))

\int \frac{1}{36 x ^{2} - 49} d x

\frac{1}{6} ln (\frac{36 x ^{2} - 49 + 6 x}{7}) + C

Solution steps

\int \frac{1}{36 x ^{2} - 49} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{6} d u

= \int \frac{sec ( u )}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{6} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{6}{7} x)

= \frac{1}{6} ln tan (arcsec (\frac{6}{7} x)) + sec (arcsec (\frac{6}{7} x))

Simplify

\frac{1}{6} ln tan (arcsec (\frac{6}{7} x)) + sec (arcsec (\frac{6}{7} x)) : \frac{1}{6} ln (\frac{36 x ^{2} - 49 + 6 x}{7})

= \frac{1}{6} ln (\frac{36 x ^{2} - 49 + 6 x}{7})

Add a constant to the solution

= \frac{1}{6} ln (\frac{36 x ^{2} - 49 + 6 x}{7}) + C

\frac{d}{d x} (\frac{4 x}{( x ^{2} + 2 ) ^{2}})

\frac{\partial}{\partial z} (x^{2} + y^{2} + z^{2})

x \to 0 + lim (x^{s i n (2 x)})

y = 6 + 5 x^{2} - 2 x^{3}

\frac{d}{d x} (sin (2 x) cos (3 x))

What is the integral of 1/(sqrt(36x^2-49)) ?
The integral of 1/(sqrt(36x^2-49)) is 1/6 ln((|sqrt(36x^2-49)+6x|)/7)+C