Find the Function f(x)=x/(x^2+1)

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

The function $F (x)$ can be found by evaluating the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Let $u = x^{2} + 1$ . Then $d u = 2 x d x$ , so $\frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

Let $u = x^{2} + 1$ . Find $\frac{d u}{d x}$ .

Differentiate $x^{2} + 1$ .

$\frac{d}{d x} [x^{2} + 1]$

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$2 x + \frac{d}{d x} [1]$

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$2 x + 0$

Add $2 x$ and $0$ .

$2 x$

Rewrite the problem using $u$ and $d u$ .

$\int \frac{1}{u} \cdot \frac{1}{2} d u$

Simplify.

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$\int \frac{1}{u \cdot 2} d u$

Move $2$ to the left of $u$ .

$\int \frac{1}{2 u} d u$

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int \frac{1}{u} d u$

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$\frac{1}{2} (\ln (| u |) + C)$

Simplify.

$\frac{1}{2} \ln (| u |) + C$

Replace all occurrences of $u$ with $x^{2} + 1$ .

$\frac{1}{2} \ln (| x^{2} + 1 |) + C$

The function $F$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

$F (x) = \frac{1}{2} \cdot \ln (| x^{2} + 1 |) + C$