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

The integral of (3x)/((6-x^2)^3) is 3/(4(6-x^2)^2)+C

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

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

\frac{3}{4 ( 6 - x ^{2} ) ^{2}} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

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

Substitute back

u = 6 - x^{2}

= 3 (- \frac{1}{2} (- \frac{1}{2 ( 6 - x ^{2} ) ^{2}}))

Simplify

3 (- \frac{1}{2} (- \frac{1}{2 ( 6 - x ^{2} ) ^{2}})) : \frac{3}{4 ( 6 - x ^{2} ) ^{2}}

= \frac{3}{4 ( 6 - x ^{2} ) ^{2}}

Add a constant to the solution

= \frac{3}{4 ( 6 - x ^{2} ) ^{2}} + C

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

\frac{d ^{2}}{d t ^{2}} (cos^{2} (t))

\frac{d}{d x} (\frac{ln ( 3 x )}{x})

3 y^{^{''}} + 13 y^{^{'}} + 4 = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (3 (7 - 3 x^{3})^{3})

What is the integral of (3x)/((6-x^2)^3) ?
The integral of (3x)/((6-x^2)^3) is 3/(4(6-x^2)^2)+C