What is the integral of (x^2)/((1-16x^2)^{3/2)} ?

The integral of (x^2)/((1-16x^2)^{3/2)} is 1/64 (-arcsin(4x)+(4x)/((1-16x^2)^{1/2)})+C

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integral of (x^2)/((1-16x^2)^{3/2)}

\int \frac{x ^{2}}{( 1 - 16 x ^{2} ) ^{\frac{3}{2}}} d x

\frac{1}{64} (- arcsin (4 x) + \frac{4 x}{( 1 - 16 x ^{2} ) ^{\frac{1}{2}}}) + C

Solution steps

\int \frac{x ^{2}}{( 1 - 16 x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{tan ^{2} ( u )}{64} d u

= \int \frac{tan ^{2} ( u )}{64} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{64} \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

= \frac{1}{64} (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= \frac{1}{64} (- u + tan (u))

Substitute back

u = arcsin (4 x)

= \frac{1}{64} (- arcsin (4 x) + tan (arcsin (4 x)))

Simplify

\frac{1}{64} (- arcsin (4 x) + tan (arcsin (4 x))) : \frac{1}{64} (- arcsin (4 x) + \frac{4 x}{( 1 - 16 x ^{2} ) ^{\frac{1}{2}}})

= \frac{1}{64} (- arcsin (4 x) + \frac{4 x}{( 1 - 16 x ^{2} ) ^{\frac{1}{2}}})

Add a constant to the solution

= \frac{1}{64} (- arcsin (4 x) + \frac{4 x}{( 1 - 16 x ^{2} ) ^{\frac{1}{2}}}) + C

x \to 5 - lim (5 - x)

\int \frac{x ^{5} + 7}{x ^{2}} d x

h^{^{''}} = 40 + 4 t

\frac{d}{d x} (\frac{1}{4 + x ^{2}})

x \to \infty lim (sin (\frac{( 2 x - 1 ) π}{2}))

What is the integral of (x^2)/((1-16x^2)^{3/2)} ?
The integral of (x^2)/((1-16x^2)^{3/2)} is 1/64 (-arcsin(4x)+(4x)/((1-16x^2)^{1/2)})+C