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

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

integral of x/((x^2+6)^2)

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

- \frac{1}{2 ( x ^{2} + 6 )} + C

Solution steps

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

Apply u-substitution

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

Substitute back

u = x^{2} + 6

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (3 e^{s i n (x + 4 c t)})

\frac{( x + 1 ) d y}{d x} + (x + 2) y = 2 x e^{- x}

\int \frac{c}{x} d x

x \to 0.1 lim (\frac{e ^{5 x} - 1}{x})

4 y^{^{''}} + 36 y^{^{'}} + 81 y = 0

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