What is the integral of x/((x^2-4)^3) ?

The integral of x/((x^2-4)^3) is -1/(4(x^2-4)^2)+C

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integral of x/((x^2-4)^3)

\int \frac{x}{( x ^{2} - 4 ) ^{3}} d x

- \frac{1}{4 ( x ^{2} - 4 ) ^{2}} + C

Solution steps

\int \frac{x}{( x ^{2} - 4 ) ^{3}} d x

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} - 4

= \frac{1}{2} (- \frac{1}{2 ( x ^{2} - 4 ) ^{2}})

Simplify

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

= - \frac{1}{4 ( x ^{2} - 4 ) ^{2}}

Add a constant to the solution

= - \frac{1}{4 ( x ^{2} - 4 ) ^{2}} + C

\frac{d}{d x} (\frac{a x ^{2}}{ln ( \frac{x}{4} )})

y^{^{'}} = \frac{x - y}{x}

e^{- (x + 2)} - sin (x^{2}) + 1

\frac{x ^{2}}{4} + x + 1

\int \frac{1}{sin ( x ) - 2 csc ( x )} d x

What is the integral of x/((x^2-4)^3) ?
The integral of x/((x^2-4)^3) is -1/(4(x^2-4)^2)+C