What is the integral of sqrt(16x^2-1) ?

The integral of sqrt(16x^2-1) is 1/8 (4xsqrt(16x^2-1)-ln|4x+sqrt(16x^2-1)|)+C

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integral of sqrt(16x^2-1)

\int 16 x^{2} - 1 d x

\frac{1}{8} (4 x 16 x^{2} - 1 - ln 4 x + 16 x^{2} - 1) + C

Solution steps

\int 16 x^{2} - 1 d x

Apply Trigonometric Substitution

: \int \frac{1}{4} tan^{2} (u) sec (u) d u

= \int \frac{1}{4} tan^{2} (u) sec (u) d u

Take the constant out:

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

= \frac{1}{4} \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

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

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= \frac{1}{4} \cdot \int - sec (u) + sec^{3} (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}{4} (- \int sec (u) d u + \int sec^{3} (u) d u)

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

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= \frac{1}{4} (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (4 x)

= \frac{1}{4} (- ln ∣ tan (arcsec (4 x)) + sec (arcsec (4 x)) ∣ + \frac{1}{2} sec (arcsec (4 x)) tan (arcsec (4 x)) + \frac{1}{2} ln ∣ tan (arcsec (4 x)) + sec (arcsec (4 x)) ∣)

Simplify

\frac{1}{4} (- ln ∣ tan (arcsec (4 x)) + sec (arcsec (4 x)) ∣ + \frac{1}{2} sec (arcsec (4 x)) tan (arcsec (4 x)) + \frac{1}{2} ln ∣ tan (arcsec (4 x)) + sec (arcsec (4 x)) ∣) : \frac{1}{8} (4 x 16 x^{2} - 1 - ln 4 x + 16 x^{2} - 1)

= \frac{1}{8} (4 x 16 x^{2} - 1 - ln 4 x + 16 x^{2} - 1)

Add a constant to the solution

= \frac{1}{8} (4 x 16 x^{2} - 1 - ln 4 x + 16 x^{2} - 1) + C

\int \frac{1}{v ln ( v )} d v

y = 3 sin (x), y = e^{7 x}, x = 0, x = \frac{π}{2}

\frac{\partial}{\partial x} (y^{5} sin (3 x))

\int_{0}^{2} 6^{x} d x

x \to 0 lim (\frac{9}{x} - \frac{9}{x ^{2} + x})

What is the integral of sqrt(16x^2-1) ?
The integral of sqrt(16x^2-1) is 1/8 (4xsqrt(16x^2-1)-ln|4x+sqrt(16x^2-1)|)+C