What is the integral of 1/(sqrt((1-u^2)^3)) ?

The integral of 1/(sqrt((1-u^2)^3)) is u/(sqrt(1-u^2))+C

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integral of 1/(sqrt((1-u^2)^3))

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

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

Solution steps

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

(1 - u^{2})^{3} = (1 - u^{2})^{\frac{3}{2}},

assuming

(1 - u^{2}) \geq 0

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

Apply Trigonometric Substitution

: \int sec^{2} (v) d v

= \int sec^{2} (v) d v

Use the common integral:

\int sec^{2} (v) d v = tan (v)

= tan (v)

Substitute back

v = arcsin (u)

= tan (arcsin (u))

Use the following identity:

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (\frac{2 x y ^{2}}{2 + x ^{2} y ^{2}})

\int \frac{82 x + 8}{( 9 x + 1 ) ( x - 1 )} d x

x \to + 6 lim (x^{2})

\int 4 + 9 x d x

y = \frac{2 x}{x ^{2} + 1}, (1, 1)

What is the integral of 1/(sqrt((1-u^2)^3)) ?
The integral of 1/(sqrt((1-u^2)^3)) is u/(sqrt(1-u^2))+C