What is the integral of x/((x^2+8)^2) ?

The integral of x/((x^2+8)^2) is -1/(2(x^2+8))+C

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integral of x/((x^2+8)^2)

\int \frac{x}{( x ^{2} + 8 ) ^{2}} d x

- \frac{1}{2 ( x ^{2} + 8 )} + C

Solution steps

\int \frac{x}{( x ^{2} + 8 ) ^{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = x^{2} + 8

= \frac{1}{2} (- \frac{1}{x ^{2} + 8})

Simplify

\frac{1}{2} (- \frac{1}{x ^{2} + 8}) : - \frac{1}{2 ( x ^{2} + 8 )}

= - \frac{1}{2 ( x ^{2} + 8 )}

Add a constant to the solution

= - \frac{1}{2 ( x ^{2} + 8 )} + C

f (x) = x^{- 12}

\frac{d}{d x} (4 x^{3} + 4 x^{2} - 8 x)

\frac{d}{d x} (ln (x + x^{2} - 11))

f (x) = x + 3

\frac{d y}{d x} = y (x y^{6} + 4)

What is the integral of x/((x^2+8)^2) ?
The integral of x/((x^2+8)^2) is -1/(2(x^2+8))+C