What is the integral of (sqrt(9x^2-16))/x ?

The integral of (sqrt(9x^2-16))/x is -4arcsec(3/4 x)+sqrt(9x^2-16)+C

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integral of (sqrt(9x^2-16))/x

\int \frac{9 x ^{2} - 16}{x} d x

- 4 arcsec (\frac{3}{4} x) + 9 x^{2} - 16 + C

Solution steps

\int \frac{9 x ^{2} - 16}{x} d x

Apply Trigonometric Substitution

: \int 4 tan^{2} (u) d u

= \int 4 tan^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 4 \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

= 4 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 4 (- u + tan (u))

Substitute back

u = arcsec (\frac{3}{4} x)

= 4 (- arcsec (\frac{3}{4} x) + tan (arcsec (\frac{3}{4} x)))

Simplify

4 (- arcsec (\frac{3}{4} x) + tan (arcsec (\frac{3}{4} x))) : - 4 arcsec (\frac{3}{4} x) + 9 x^{2} - 16

= - 4 arcsec (\frac{3}{4} x) + 9 x^{2} - 16

Add a constant to the solution

= - 4 arcsec (\frac{3}{4} x) + 9 x^{2} - 16 + C

f (x) = 4^{(x^{9})}

n = 0 \sum \infty \frac{1}{1 + n}

\frac{d}{d x} (x (x^{3} + x^{2} - 3 x))

\int \frac{2 x + 3}{x ^{3} + 3 x} d x

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

What is the integral of (sqrt(9x^2-16))/x ?
The integral of (sqrt(9x^2-16))/x is -4arcsec(3/4 x)+sqrt(9x^2-16)+C