What is the integral of (sqrt(x^2-25))/(5x) ?

The integral of (sqrt(x^2-25))/(5x) is -arcsec(1/5 x)+1/5 sqrt(x^2-25)+C

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integral of (sqrt(x^2-25))/(5x)

\int \frac{x ^{2} - 25}{5 x} d x

- arcsec (\frac{1}{5} x) + \frac{1}{5} x^{2} - 25 + C

Solution steps

\int \frac{x ^{2} - 25}{5 x} d x

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{x ^{2} - 25}{x} d x

Apply Trigonometric Substitution

: \int 5 tan^{2} (u) d u

= \frac{1}{5} \cdot \int 5 tan^{2} (u) d u

Take the constant out:

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

= \frac{1}{5} \cdot 5 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{5} \cdot 5 \cdot \int - 1 + sec^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{5} \cdot 5 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= \frac{1}{5} \cdot 5 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{5} x)

= \frac{1}{5} \cdot 5 (- arcsec (\frac{1}{5} x) + tan (arcsec (\frac{1}{5} x)))

Simplify

\frac{1}{5} \cdot 5 (- arcsec (\frac{1}{5} x) + tan (arcsec (\frac{1}{5} x))) : - arcsec (\frac{1}{5} x) + \frac{1}{5} x^{2} - 25

= - arcsec (\frac{1}{5} x) + \frac{1}{5} x^{2} - 25

Add a constant to the solution

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

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