What is the integral of (sqrt(36x^2-1))/x ?

The integral of (sqrt(36x^2-1))/x is -arctan(sqrt(36x^2-1))+sqrt(36x^2-1)+C

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integral of (sqrt(36x^2-1))/x

\int \frac{36 x ^{2} - 1}{x} d x

- arctan (36 x^{2} - 1) + 36 x^{2} - 1 + C

Solution steps

\int \frac{36 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 = 36 x^{2} - 1

= - arctan (36 x^{2} - 1) + 36 x^{2} - 1

Add a constant to the solution

= - arctan (36 x^{2} - 1) + 36 x^{2} - 1 + C

(3 x - 2)^{2} (3 - x^{2})^{2}

\int (16 - x^{2}) d x

x \to 3 - lim (\frac{2}{e ^{3 - x}})

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

y = g (x), at x = 2

What is the integral of (sqrt(36x^2-1))/x ?
The integral of (sqrt(36x^2-1))/x is -arctan(sqrt(36x^2-1))+sqrt(36x^2-1)+C