What is the integral of x/((x^2-1)[ln(x^2-1)]^2) ?

The integral of x/((x^2-1)[ln(x^2-1)]^2) is -1/(2ln(x^2-1))+C

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integral of x/((x^2-1)[ln(x^2-1)]^2)

\int \frac{x}{( x ^{2} - 1 ) [ ln ( x ^{2} - 1 ) ] ^{2}} d x

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

Solution steps

\int \frac{x}{( x ^{2} - 1 ) [ ln ( x ^{2} - 1 ) ] ^{2}} d x

Apply u-substitution

: \int \frac{1}{2 u ln ^{2} ( u )} d u

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{2} (- \frac{1}{v})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} = 5 y^{2} sin (x)

\frac{d}{d x} (x^{c o t (x)})

\frac{\partial}{\partial y} (sin (\frac{x}{y}))

\int x^{2} - 13 d x

\frac{d}{d x} (- \frac{20000}{1 + 0.2 x})

What is the integral of x/((x^2-1)[ln(x^2-1)]^2) ?
The integral of x/((x^2-1)[ln(x^2-1)]^2) is -1/(2ln(x^2-1))+C