What is the integral of $\frac{x}{1 + x^{2}}$ $x/(1+x^2)$ ?

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What is the integral of $\frac{x}{1 + x^{2}}$ $x/(1+x^2)$ ?

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Answer 1 · 2016-11-21T11:10:25

$\int \frac{x}{x^{2} + 1} d x = \frac{1}{2} ln (x^{2} + 1) + C$ $intx/(x^2+1)dx=1/2ln(x^2+1)+C$

Explanation:

Let $u (x) = 1 + x^{2}$ $u(x)=1+x^2" "$ then $d u (x) = 2 x d x$ $" "du(x) =2xdx$
$" "$
$\frac{d (u (x))}{2} = x d x$ $color(blue)((d(u(x)))/2=xdx)$
$" "$
Start solving the integral.
$" "$
$\int \frac{x}{x^{2} + 1} d x$ $intx/(x^2+1)dx$
$" "$
$= \int \frac{d (u (x))}{2 u (x)}$ $=intcolor(blue)((d(u(x)))/(2u(x))$
$" "$
$= \frac{1}{2} \int \frac{d u (x)}{u (x)}$ $=1/2int(du(x))/(u(x))$
$" "$
$= \frac{1}{2} ln | u (x) | + C$ $=1/2lnabs(u(x))+C$
$" "$
$= \frac{1}{2} ln ∣ ∣ x^{2} + 1 ∣ ∣ + C$ $=1/2lnabs(x^2+1)+C$
$" "$
Because $x^{2} + 1 > 0 t h e n ∣ ∣ x^{2} + 1 ∣ ∣ = x^{2} + 1$ $x^2+1>0 " " then " " abs(x^2+1)=x^2+1$
$" "$
Therefore,
$" "$
$\int \frac{x}{x^{2} + 1} d x = \frac{1}{2} ln (x^{2} + 1) + C$ $intx/(x^2+1)dx=1/2ln(x^2+1)+C$

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