What is the integral of x/(sqrt(x^2)\sqrt{x^2-1)} ?

The integral of x/(sqrt(x^2)\sqrt{x^2-1)} is ln|x+sqrt(x^2-1)|+C

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integral of x/(sqrt(x^2)\sqrt{x^2-1)}

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

ln x + x^{2} - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u u - 1} d u

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u u - 1} d u

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = ln v^{2} + 1 + v

= \frac{1}{2} \cdot 2 ln v^{2} + 1 + v

Substitute back

= \frac{1}{2} \cdot 2 ln (x^{2} - 1)^{2} + 1 + x^{2} - 1

Simplify

\frac{1}{2} \cdot 2 ln (x^{2} - 1)^{2} + 1 + x^{2} - 1 : ln x + x^{2} - 1

= ln x + x^{2} - 1

Add a constant to the solution

= ln x + x^{2} - 1 + C

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What is the integral of x/(sqrt(x^2)\sqrt{x^2-1)} ?
The integral of x/(sqrt(x^2)\sqrt{x^2-1)} is ln|x+sqrt(x^2-1)|+C