What is the integral of (6sqrt(x^2-1))/x ?

The integral of (6sqrt(x^2-1))/x is 6(-arctan(sqrt(x^2-1))+sqrt(x^2-1))+C

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integral of (6sqrt(x^2-1))/x

\int \frac{6 x ^{2} - 1}{x} d x

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

Solution steps

\int \frac{6 x ^{2} - 1}{x} d x

Take the constant out:

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

= 6 \cdot \int \frac{x ^{2} - 1}{x} d x

Apply u-substitution

= 6 \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

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

= 6 \cdot \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

= 6 (- \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

= 6 (- arctan (u) + u)

Substitute back

u = x^{2} - 1

= 6 (- arctan (x^{2} - 1) + x^{2} - 1)

Add a constant to the solution

= 6 (- arctan (x^{2} - 1) + x^{2} - 1) + C

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What is the integral of (6sqrt(x^2-1))/x ?
The integral of (6sqrt(x^2-1))/x is 6(-arctan(sqrt(x^2-1))+sqrt(x^2-1))+C