What is the integral of x/((3-x^2)^6) ?

The integral of x/((3-x^2)^6) is 1/(10(3-x^2)^5)+C

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integral of x/((3-x^2)^6)

\int \frac{x}{( 3 - x ^{2} ) ^{6}} d x

\frac{1}{10 ( 3 - x ^{2} ) ^{5}} + C

Solution steps

\int \frac{x}{( 3 - x ^{2} ) ^{6}} d x

Apply u-substitution

: \int - \frac{1}{2 u ^{6}} d u

= \int - \frac{1}{2 u ^{6}} 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 ^{6}} d u

Apply the Power Rule

: - \frac{1}{5 u ^{5}}

= - \frac{1}{2} (- \frac{1}{5 u ^{5}})

Substitute back

u = 3 - x^{2}

= - \frac{1}{2} (- \frac{1}{5 ( 3 - x ^{2} ) ^{5}})

Simplify

- \frac{1}{2} (- \frac{1}{5 ( 3 - x ^{2} ) ^{5}}) : \frac{1}{10 ( 3 - x ^{2} ) ^{5}}

= \frac{1}{10 ( 3 - x ^{2} ) ^{5}}

Add a constant to the solution

= \frac{1}{10 ( 3 - x ^{2} ) ^{5}} + C

\frac{\partial}{\partial x} (5 + x ln (x y - 5))

y^{^{'}} - x y^{^{'}} = 2 x y

\int (y^{2} e^{x y} + cos (x)) d x

5 x - 4

\int \frac{y}{y + 6} d y

What is the integral of x/((3-x^2)^6) ?
The integral of x/((3-x^2)^6) is 1/(10(3-x^2)^5)+C