What is the integral of x/(sqrt((36-x^2)^3)) ?

The integral of x/(sqrt((36-x^2)^3)) is 1/(sqrt(36-x^2))+C

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integral of x/(sqrt((36-x^2)^3))

\int \frac{x}{( 36 - x ^{2} ) ^{3}} d x

\frac{1}{36 - x ^{2}} + C

Solution steps

\int \frac{x}{( 36 - x ^{2} ) ^{3}} d x

(36 - x^{2})^{3} = (36 - x^{2})^{\frac{3}{2}},

assuming

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

= \int \frac{x}{( 36 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int - \frac{1}{2 u ^{\frac{3}{2}}} d u

= \int - \frac{1}{2 u ^{\frac{3}{2}}} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Power Rule

: - \frac{2}{u}

= - \frac{1}{2} (- \frac{2}{u})

Substitute back

u = 36 - x^{2}

= - \frac{1}{2} (- \frac{2}{36 - x ^{2}})

Simplify

- \frac{1}{2} (- \frac{2}{36 - x ^{2}}) : \frac{1}{36 - x ^{2}}

= \frac{1}{36 - x ^{2}}

Add a constant to the solution

= \frac{1}{36 - x ^{2}} + C

x \to \infty lim (1 + x^{2})

(e^{s i n (x)})^{^{'}}

\int_{1}^{4} e^{x} d x

y = (x^{2} + 1)^{5} (1 - x^{2})^{5}

f (x) = x^{3} - 2 x^{2} + 5

What is the integral of x/(sqrt((36-x^2)^3)) ?
The integral of x/(sqrt((36-x^2)^3)) is 1/(sqrt(36-x^2))+C