What is the integral of (x^2)/((9-9x^2)^{3/2)} ?

The integral of (x^2)/((9-9x^2)^{3/2)} is 1/27 (-arcsin(x)+x/((1-x^2)^{1/2)})+C

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integral of (x^2)/((9-9x^2)^{3/2)}

\int \frac{x ^{2}}{( 9 - 9 x ^{2} ) ^{\frac{3}{2}}} d x

\frac{1}{27} (- arcsin (x) + \frac{x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}) + C

Solution steps

\int \frac{x ^{2}}{( 9 - 9 x ^{2} ) ^{\frac{3}{2}}} d x

9 - 9 x^{2} = 9 (1 - \frac{9 x ^{2}}{9})

= \int \frac{x ^{2}}{( 9 ( 1 - \frac{9 x ^{2}}{9} ) ) ^{\frac{3}{2}}} d x

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{9 x ^{2}}{9} = (x)^{2}

= \int \frac{x ^{2}}{( 9 ( 1 - x ^{2} ) ) ^{\frac{3}{2}}} d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

(9 (1 - (x)^{2}))^{\frac{3}{2}} = (9)^{\frac{3}{2}} ((1 - (x)^{2}))^{\frac{3}{2}}

= \int \frac{x ^{2}}{9 ^{\frac{3}{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

= \frac{1}{9 ^{\frac{3}{2}}} \cdot \int \frac{x ^{2}}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

9^{\frac{3}{2}} = 27

= \frac{1}{27} \cdot \int \frac{x ^{2}}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

= \frac{1}{27} \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{27} \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}{27} (- \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}{27} (- u + tan (u))

Substitute back

u = arcsin (x)

= \frac{1}{27} (- arcsin (x) + tan (arcsin (x)))

Simplify

\frac{1}{27} (- arcsin (x) + tan (arcsin (x))) : \frac{1}{27} (- arcsin (x) + \frac{x}{( 1 - x ^{2} ) ^{\frac{1}{2}}})

= \frac{1}{27} (- arcsin (x) + \frac{x}{( 1 - x ^{2} ) ^{\frac{1}{2}}})

Add a constant to the solution

= \frac{1}{27} (- arcsin (x) + \frac{x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}) + C

What is the integral of (x^2)/((9-9x^2)^{3/2)} ?
The integral of (x^2)/((9-9x^2)^{3/2)} is 1/27 (-arcsin(x)+x/((1-x^2)^{1/2)})+C