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

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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

= \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} + 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{d y}{d t} - 4 y = 2 e^{5 t}

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

\frac{\frac{1}{s + 1} - 1}{s ^{2} + 1}

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

\frac{1}{x x}

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