What is the integral of 1/(sqrt(25x^2-49)) ?

The integral of 1/(sqrt(25x^2-49)) is 1/5 ln((|sqrt(25x^2-49)+5x|)/7)+C

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integral of 1/(sqrt(25x^2-49))

\int \frac{1}{25 x ^{2} - 49} d x

\frac{1}{5} ln (\frac{25 x ^{2} - 49 + 5 x}{7}) + C

Solution steps

\int \frac{1}{25 x ^{2} - 49} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{5} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = arcsec (\frac{5}{7} x)

= \frac{1}{5} ln tan (arcsec (\frac{5}{7} x)) + sec (arcsec (\frac{5}{7} x))

Simplify

\frac{1}{5} ln tan (arcsec (\frac{5}{7} x)) + sec (arcsec (\frac{5}{7} x)) : \frac{1}{5} ln (\frac{25 x ^{2} - 49 + 5 x}{7})

= \frac{1}{5} ln (\frac{25 x ^{2} - 49 + 5 x}{7})

Add a constant to the solution

= \frac{1}{5} ln (\frac{25 x ^{2} - 49 + 5 x}{7}) + C

\int \frac{6 - x}{4 x ^{2} - 12 x + 7} d x

t y^{^{'}} + 3 y = t^{2} - t + 7

x \to 0 lim (\frac{1}{x})

x \to 2 lim (x^{3} - 4 x)

f (x) = x^{2} - 5 x

What is the integral of 1/(sqrt(25x^2-49)) ?
The integral of 1/(sqrt(25x^2-49)) is 1/5 ln((|sqrt(25x^2-49)+5x|)/7)+C