What is the integral of (x^2)/(sqrt((16-x^2)^3)) ?

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

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integral of (x^2)/(sqrt((16-x^2)^3))

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

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

Solution steps

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

(16 - x^{2})^{3} = (16 - x^{2})^{\frac{3}{2}},

assuming

(16 - x^{2}) \geq 0

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \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

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

\int 1 d u = u

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

= - u + tan (u)

Substitute back

u = arcsin (\frac{1}{4} x)

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

Simplify

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

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

Add a constant to the solution

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

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