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

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

integral of x/((x^2+4)^2)

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

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

Solution steps

\int \frac{x}{( x ^{2} + 4 ) ^{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} + 4

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

Simplify

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

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

Add a constant to the solution

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

d er i v a t i v e - \frac{11}{s ^{4}}

n = 1 \sum \infty \frac{n x ^{n}}{5 ^{n}}

t an g e n t f (x) = \frac{x ^{3} + 5}{x ^{2} + 1}, at x = - 1

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

\frac{d}{d x} ((x + e^{x}) (3 - x))

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