What is the integral of 1/(sqrt(49x^2-64)) ?

The integral of 1/(sqrt(49x^2-64)) is 1/7 ln((|sqrt(49x^2-64)+7x|)/8)+C

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integral of 1/(sqrt(49x^2-64))

\int \frac{1}{49 x ^{2} - 64} d x

\frac{1}{7} ln (\frac{49 x ^{2} - 64 + 7 x}{8}) + C

Solution steps

\int \frac{1}{49 x ^{2} - 64} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{7} d u

= \int \frac{sec ( u )}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{7} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{7}{8} x)

= \frac{1}{7} ln tan (arcsec (\frac{7}{8} x)) + sec (arcsec (\frac{7}{8} x))

Simplify

\frac{1}{7} ln tan (arcsec (\frac{7}{8} x)) + sec (arcsec (\frac{7}{8} x)) : \frac{1}{7} ln (\frac{49 x ^{2} - 64 + 7 x}{8})

= \frac{1}{7} ln (\frac{49 x ^{2} - 64 + 7 x}{8})

Add a constant to the solution

= \frac{1}{7} ln (\frac{49 x ^{2} - 64 + 7 x}{8}) + C

\frac{d}{d x} (- b sin (x))

\frac{d}{d t} (\frac{t}{( t - 1 ) ^{2}})

n = 0 \sum \infty \frac{n}{n ^{3} + 4}

x^{6} + y^{2} = 2, (1, - 1)

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

What is the integral of 1/(sqrt(49x^2-64)) ?
The integral of 1/(sqrt(49x^2-64)) is 1/7 ln((|sqrt(49x^2-64)+7x|)/8)+C