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

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Mia Share
Nov 21, 2016

$\int \frac{x}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} \ln \left({x}^{2} + 1\right) + C$

#### Explanation:

Let $u \left(x\right) = 1 + {x}^{2} \text{ }$ then $\text{ } \mathrm{du} \left(x\right) = 2 x \mathrm{dx}$
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$\textcolor{b l u e}{\frac{d \left(u \left(x\right)\right)}{2} = x \mathrm{dx}}$
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Start solving the integral.
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$\int \frac{x}{{x}^{2} + 1} \mathrm{dx}$
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=intcolor(blue)((d(u(x)))/(2u(x))
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$= \frac{1}{2} \int \frac{\mathrm{du} \left(x\right)}{u \left(x\right)}$
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$= \frac{1}{2} \ln \left\mid u \left(x\right) \right\mid + C$
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$= \frac{1}{2} \ln \left\mid {x}^{2} + 1 \right\mid + C$
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Because ${x}^{2} + 1 > 0 \text{ " then " } \left\mid {x}^{2} + 1 \right\mid = {x}^{2} + 1$
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Therefore,
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$\int \frac{x}{{x}^{2} + 1} \mathrm{dx} = \frac{1}{2} \ln \left({x}^{2} + 1\right) + C$

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