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

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

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integral of (sqrt(a^2x^2-a^2))/x

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

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

Solution steps

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

a^{2} x^{2} - a^{2} = a^{2} x^{2} - 1

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

Take the constant out:

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

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

Apply u-substitution

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

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

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

= a^{2} \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

= a^{2} (- \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

= a^{2} (- arctan (u) + u)

Substitute back

u = x^{2} - 1

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

Apply radical rule:

n a^{n} = a,

assuming

a \geq 0

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

Add a constant to the solution

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

y^{^{'}} = - \frac{2 t ( 1 + y ^{2} )}{y}

\int 6 x^{2} - 2 x + 8 d x

\frac{2}{x}

x \to 2 lim (x - 3)

\frac{d}{d v} (2 uv)

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