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

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

Popular >

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

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

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

Solution steps

\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{1}{2}}}

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

Add a constant to the solution

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

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

\frac{d}{d x} (e^{s i n (3 x^{2})})

\int x 2 x - 3 d x

\int 9 tan^{5} (x) d x

e^{- (\frac{1}{x + 1})}

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