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0-x-1-1-x-2-dx-




Question Number 62937 by Prithwish sen last updated on 27/Jun/19
∫_0 ^(  x) (1/(1+x^2 )) dx
$$\int_{\mathrm{0}} ^{\:\:\mathrm{x}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$
Commented by mathmax by abdo last updated on 27/Jun/19
∫_0 ^x   (dt/(1+t^2 )) =[arctan(t)]_0 ^x  =arctanx .
$$\int_{\mathrm{0}} ^{{x}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\left[{arctan}\left({t}\right)\right]_{\mathrm{0}} ^{{x}} \:={arctanx}\:. \\ $$
Commented by Prithwish sen last updated on 27/Jun/19
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Prithwish sen last updated on 27/Jun/19
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by peter frank last updated on 27/Jun/19
let x=tan θ→θ=tan^(−1) x  dx=sec^2 θdθ  ∫((sec^2 θdθ)/(sec^2 θ))  θ+c  tan^(−1) x+c
$${let}\:{x}=\mathrm{tan}\:\theta\rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$${dx}=\mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta \\ $$$$\int\frac{\mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta}{\mathrm{sec}\:^{\mathrm{2}} \theta} \\ $$$$\theta+{c} \\ $$$$\mathrm{tan}^{−\mathrm{1}} {x}+{c} \\ $$$$ \\ $$
Commented by Prithwish sen last updated on 27/Jun/19
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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