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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Use the common integral:

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

= tan (u)

Substitute back

u = arcsin (x)

= tan (arcsin (x))

Use the following identity:

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

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

Add a constant to the solution

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

\int_{0}^{2} x e^{2 x} d x

\int_{1}^{\infty} sin^{2} (x) d x

\frac{d}{d x} (cot (x^{4}))

\frac{d}{d x} (\frac{x ^{2}}{2} - \frac{x ^{7}}{7})

(2 x y^{2} + y) d x + (2 x^{2} y + x) d y = 0

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