What is the integral of x/((x^2+1)^2) ?

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

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integral of x/((x^2+1)^2)

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{x x}

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

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

\frac{\partial}{\partial x} (sin (x y) + x + y)

\int 1 - sin (2 x) d x

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