What is the integral of y/((y^2-1)^{3/2)} ?

The integral of y/((y^2-1)^{3/2)} is -1/((y^2-1)^{1/2)}+C

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integral of y/((y^2-1)^{3/2)}

\int \frac{y}{( y ^{2} - 1 ) ^{\frac{3}{2}}} d y

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

Solution steps

\int \frac{y}{( y ^{2} - 1 ) ^{\frac{3}{2}}} d y

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{3}{2}}} d u

= \int \frac{1}{2 u ^{\frac{3}{2}}} 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 ^{\frac{3}{2}}} d u

Apply the Power Rule

: - \frac{2}{u ^{\frac{1}{2}}}

= \frac{1}{2} (- \frac{2}{u ^{\frac{1}{2}}})

Substitute back

u = y^{2} - 1

= \frac{1}{2} (- \frac{2}{( y ^{2} - 1 ) ^{\frac{1}{2}}})

Simplify

\frac{1}{2} (- \frac{2}{( y ^{2} - 1 ) ^{\frac{1}{2}}}) : - \frac{1}{( y ^{2} - 1 ) ^{\frac{1}{2}}}

= - \frac{1}{( y ^{2} - 1 ) ^{\frac{1}{2}}}

Add a constant to the solution

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

(ln (x + 1))^{^{'}}

x^{2} + 1, 0, 1

x \to \infty lim (1.0 1^{x})

(x^{2} - x + 1) y^{^{''}} - (x^{2} + x) y^{^{'}} + (x + 1) y = 0

y = 2187 x, y = 81 x^{2}

What is the integral of y/((y^2-1)^{3/2)} ?
The integral of y/((y^2-1)^{3/2)} is -1/((y^2-1)^{1/2)}+C