What is the integral of (sqrt(64x^2-1))/x ?

The integral of (sqrt(64x^2-1))/x is -arctan(sqrt(64x^2-1))+sqrt(64x^2-1)+C

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integral of (sqrt(64x^2-1))/x

\int \frac{64 x ^{2} - 1}{x} d x

- arctan (64 x^{2} - 1) + 64 x^{2} - 1 + C

Solution steps

\int \frac{64 x ^{2} - 1}{x} d x

Apply u-substitution

: \int \frac{u ^{2}}{u ^{2} + 1} d u

= \int \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

= \int - \frac{1}{u ^{2} + 1} + 1 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

= - \int \frac{1}{u ^{2} + 1} d u + \int 1 d u

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int 1 d u = u

= - arctan (u) + u

Substitute back

u = 64 x^{2} - 1

= - arctan (64 x^{2} - 1) + 64 x^{2} - 1

Add a constant to the solution

= - arctan (64 x^{2} - 1) + 64 x^{2} - 1 + C

\frac{\partial}{\partial x} (e^{x^{2} y} + x^{2} y)

\frac{d}{d x} (- \frac{y}{x ^{2}})

\frac{\partial}{\partial x} (x^{5} + x y^{7} + 6)

f (x) = 4 cos (x), (\frac{π}{3}, 2)

What is the integral of (sqrt(64x^2-1))/x ?
The integral of (sqrt(64x^2-1))/x is -arctan(sqrt(64x^2-1))+sqrt(64x^2-1)+C