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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 bymathmax 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 byPrithwish sen last updated on 27/Jun/19

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented byPrithwish 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 byPrithwish sen last updated on 27/Jun/19

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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