What is the integral of (sqrt(4x^2-25))/x ?

The integral of (sqrt(4x^2-25))/x is -5arcsec(2/5 x)+sqrt(4x^2-25)+C

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integral of (sqrt(4x^2-25))/x

\int \frac{4 x ^{2} - 25}{x} d x

- 5 arcsec (\frac{2}{5} x) + 4 x^{2} - 25 + C

Solution steps

\int \frac{4 x ^{2} - 25}{x} d x

Apply Trigonometric Substitution

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

= \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

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

Rewrite using trig identities

= 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

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

\int 1 d u = u

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

= 5 (- u + tan (u))

Substitute back

u = arcsec (\frac{2}{5} x)

= 5 (- arcsec (\frac{2}{5} x) + tan (arcsec (\frac{2}{5} x)))

Simplify

5 (- arcsec (\frac{2}{5} x) + tan (arcsec (\frac{2}{5} x))) : - 5 arcsec (\frac{2}{5} x) + 4 x^{2} - 25

= - 5 arcsec (\frac{2}{5} x) + 4 x^{2} - 25

Add a constant to the solution

= - 5 arcsec (\frac{2}{5} x) + 4 x^{2} - 25 + C

(- 6, 1), (- 4, - 9)

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

\int \frac{2}{x ^{2} + 4} d x

y = \frac{1}{3} x^{3} + 2, y = x + 1, 0, 2

\int \frac{sin ^{3} ( x )}{( x )} d x

What is the integral of (sqrt(4x^2-25))/x ?
The integral of (sqrt(4x^2-25))/x is -5arcsec(2/5 x)+sqrt(4x^2-25)+C