What is the integral of (x^2)/((9-25x^2)^{5/2)} ?

The integral of (x^2)/((9-25x^2)^{5/2)} is (x^3)/(27(-25x^2+9)^{3/2)}+C

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integral of (x^2)/((9-25x^2)^{5/2)}

\int \frac{x ^{2}}{( 9 - 25 x ^{2} ) ^{\frac{5}{2}}} d x

\frac{x ^{3}}{27 ( - 25 x ^{2} + 9 ) ^{\frac{3}{2}}} + C

Solution steps

\int \frac{x ^{2}}{( 9 - 25 x ^{2} ) ^{\frac{5}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{sin ^{2} ( u )}{1125 cos ^{4} ( u )} d u

= \int \frac{sin ^{2} ( u )}{1125 cos ^{4} ( u )} d u

Take the constant out:

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

= \frac{1}{1125} \cdot \int \frac{sin ^{2} ( u )}{cos ^{4} ( u )} d u

Rewrite using trig identities

= \frac{1}{1125} \cdot \int \frac{tan ^{2} ( u )}{cos ^{2} ( u )} d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{1125} \cdot \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{1125} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{1125} \cdot \frac{tan ^{3} ( arcsin ( \frac{5}{3} x ) )}{3}

Simplify

\frac{1}{1125} \cdot \frac{tan ^{3} ( arcsin ( \frac{5}{3} x ) )}{3} : \frac{x ^{3}}{27 ( - 25 x ^{2} + 9 ) ^{\frac{3}{2}}}

= \frac{x ^{3}}{27 ( - 25 x ^{2} + 9 ) ^{\frac{3}{2}}}

Add a constant to the solution

= \frac{x ^{3}}{27 ( - 25 x ^{2} + 9 ) ^{\frac{3}{2}}} + C

\frac{1 - x}{1 + x}

\frac{d}{d x} (\frac{1}{5} e^{x})

x \to 2 lim (\frac{2 ^{x} - 4}{x - 2})

y = e^{x}, y = x, x = 0, x = 2

\int \frac{6 x ^{2} + 13 x + 6}{( x + 2 ) ( x + 1 ) ^{2}} d x

What is the integral of (x^2)/((9-25x^2)^{5/2)} ?
The integral of (x^2)/((9-25x^2)^{5/2)} is (x^3)/(27(-25x^2+9)^{3/2)}+C