What is the integral of (4x^2)/((1-x^2)^{3/2)} ?

The integral of (4x^2)/((1-x^2)^{3/2)} is 4(-arcsin(x)+x/((1-x^2)^{1/2)})+C

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integral of (4x^2)/((1-x^2)^{3/2)}

\int \frac{4 x ^{2}}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

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

Solution steps

\int \frac{4 x ^{2}}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{x ^{2}}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

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

= 4 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 4 \cdot \int - 1 + sec^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 4 (- u + tan (u))

Substitute back

u = arcsin (x)

= 4 (- arcsin (x) + tan (arcsin (x)))

Simplify

4 (- arcsin (x) + tan (arcsin (x))) : 4 (- arcsin (x) + \frac{x}{( 1 - x ^{2} ) ^{\frac{1}{2}}})

= 4 (- arcsin (x) + \frac{x}{( 1 - x ^{2} ) ^{\frac{1}{2}}})

Add a constant to the solution

= 4 (- arcsin (x) + \frac{x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}) + C

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What is the integral of (4x^2)/((1-x^2)^{3/2)} ?
The integral of (4x^2)/((1-x^2)^{3/2)} is 4(-arcsin(x)+x/((1-x^2)^{1/2)})+C