# Antiderivatives and definite integrals Compute the following integral: I = <msubsup>

Antiderivatives and definite integrals
Compute the following integral:
$I={\int }_{0}^{1}\left({\int }_{1}^{x}\frac{1}{1+{t}^{2}}dt\right)dx$
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Explanation:
$\begin{array}{rl}{\int }_{0}^{1}{\int }_{1}^{x}\frac{dt}{1+{t}^{2}}\cdot dx& ={\int }_{0}^{1}\left(\mathrm{arctan}\left(x\right)-\frac{\pi }{4}\right)dx\\ & =\left[x\mathrm{arctan}\left(x\right)-\frac{1}{2}\mathrm{ln}\left(1+{x}^{2}\right)-\frac{\pi }{4}x{\right]}_{0}^{1}=-\frac{1}{2}\mathrm{ln}\left(2\right)\end{array}$
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dream13rxs
Step 1
Let $f\left(t\right)=\frac{1}{1+{t}^{2}}$, $t\in \mathbb{R}$ and $F\left(x\right)={\int }_{1}^{x}\frac{1}{1+{t}^{2}}dt={\int }_{1}^{x}f\left(t\right)dt$, $t\in \mathbb{R}$
Step 2
Consequently, ${F}^{\prime }\left(x\right)=f\left(x\right)$, $x\in \mathbb{R}$ and:
$I={\int }_{0}^{1}\left({\int }_{1}^{x}\frac{1}{1+{t}^{2}}dt\right)dx={\int }_{0}^{1}F\left(x\right)dx={\int }_{0}^{1}\left(x{\right)}^{\prime }F\left(x\right)dx=$
$=\left[xF\left(x\right){\right]}_{0}^{1}-{\int }_{0}^{1}x{F}^{\prime }\left(x\right)dx=F\left(1\right)-{\int }_{0}^{1}xf\left(x\right)dx=0-{\int }_{0}^{1}\frac{x}{1+{x}^{2}}dx=$
$=-\frac{1}{2}{\int }_{0}^{1}\frac{\left(1+{x}^{2}{\right)}^{\prime }}{1+{x}^{2}}dx=-\frac{1}{2}\left[\mathrm{ln}\left(1+{x}^{2}\right){\right]}_{0}^{1}=-\frac{\mathrm{ln}2}{2}$