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Question Number 201291 by sonukgindia last updated on 03/Dec/23

Answered by aleks041103 last updated on 03/Dec/23

$$I={\int }_0^1\frac{ln(1+x)dx}{1+x^2}$$$$notice$$$$x=\frac{1-t}{1+t},dx=-\frac{2dt}{(1+t^2)}$$$$1+x=\frac{1-t}{1+t}+1=\frac{2}{1+t}$$$$1+x^2=1+\frac{1+t^2-2t}{1+t^2+2t}=\frac{2+2t^2}{1+t^2+2t}=\frac{2(1+t^2)}{{(1+t)}^2}$$$$\Rightarrow I={\int }_1^0\frac{ln\left(2\right)-ln(1+t)}{\frac{2(1+t^2)}{{(1+t)}^2}}(-\frac{2dt}{{(1+t)}^2})=$$$$={\int }_0^1\frac{ln\left(2\right)-ln(1+t)}{1+t^2}dt=ln\left(2\right){\int }_0^1\frac{dt}{1+t^2}-I$$$$\Rightarrow 2I=ln\left(2\right){\int }_0^1\frac{dt}{1+t^2}=ln\left(2\right)arctan\left(1\right)$$$$\Rightarrow I=\frac{\pi ln\left(2\right)}{8}$$

Answered by Calculusboy last updated on 04/Dec/23

$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:{\int }_0^1\frac{\mathit{I}\mathit{n}(\mathit{x}+1)}{1+{\mathit{x}}^2}\mathit{d}\mathit{x}\mathit{l}\mathit{e}\mathit{t}\mathit{x}=\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }\mathit{d}\mathit{x}={\mathit{s}\mathit{e}\mathit{c}}^2\mathit{\theta }\mathit{d}\mathit{\theta }$$$$\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathit{x}=1\mathit{\theta }=\frac{\mathit{\pi }}{4}\mathit{a}\mathit{n}\mathit{d}\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathit{x}=0\mathit{\theta }=0$$$$\mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\frac{\mathit{I}\mathit{n}(\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }+1)}{1+{\mathit{t}\mathit{a}\mathit{n}}^2\mathit{\theta }}{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{\theta }\mathit{d}\mathit{\theta }\mathit{N}\mathit{b}:1+{\mathit{t}\mathit{a}\mathit{n}}^2\mathit{\theta }={\mathit{s}\mathit{e}\mathit{c}}^2\mathit{\theta }$$$$\mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\frac{\mathit{I}\mathit{n}(1+\mathit{t}\mathit{a}\mathit{n}\mathit{\theta })}{{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{\theta }}{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{\theta }\mathit{d}\mathit{\theta }\iff \mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}(1+\mathit{t}\mathit{a}\mathit{n}\mathit{\theta })\mathit{d}\mathit{\theta }$$$$\mathit{l}\mathit{e}\mathit{t}\mathit{y}=\frac{\mathit{\pi }}{4}-\mathit{\theta }\mathit{d}\mathit{y}=-\mathit{d}\mathit{\theta }$$$$\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathit{\theta }=\frac{\mathit{\pi }}{4}\mathit{y}=0\mathit{a}\mathit{n}\mathit{d}\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathit{\theta }=0\mathit{y}=\frac{\mathit{\pi }}{4}$$$$\mathit{I}={\int }_{\frac{\mathit{\pi }}{4}}^0\mathit{I}\mathit{n}[1+\mathit{t}\mathit{a}\mathit{n}(\frac{\mathit{\pi }}{4}-\mathit{y})](-\mathit{d}\mathit{y})$$$$\mathit{c}\mathit{h}\mathit{a}\mathit{n}\mathit{g}\mathit{i}\mathit{n}\mathit{g}\mathit{o}\mathit{f}\mathit{v}\mathit{a}\mathit{r}\mathit{i}\mathit{a}\mathit{b}\mathit{l}\mathit{e}$$$$\mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}[1+\mathit{t}\mathit{a}\mathit{n}(\frac{\mathit{\pi }}{4}-\mathit{\theta })]\mathit{d}\mathit{\theta }\iff \mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}[1+\frac{\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{\pi }}{4}\right)-\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}{1+\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{\pi }}{4}\right)\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}]\mathit{d}\mathit{\theta }$$$$\mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}[1+\frac{1-\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}{1+\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}]\mathit{d}\mathit{\theta }\iff \mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}\left[\frac{1+\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }+1-\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}{1+\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}\right]\mathit{d}\mathit{\theta }\iff \mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}\left[\frac{2}{1+\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}\right]\mathit{d}\mathit{\theta }$$$$\mathit{I}={\int }_0^{\frac{\mathit{\pi }}{4}}[\mathit{I}\mathit{n}2-\mathit{I}\mathit{n}(1+\mathit{t}\mathit{a}\mathit{n}\mathit{\theta })]\mathit{d}\mathit{\theta }\iff \mathit{I}=\mathit{I}\mathit{n}\left(2\right){\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{d}\mathit{\theta }-{\int }_0^{\frac{\mathit{\pi }}{4}}\mathit{I}\mathit{n}(1+\mathit{t}\mathit{a}\mathit{n}\mathit{\theta })\mathit{d}\mathit{\theta }$$$$\mathit{I}=\mathit{I}\mathit{n}\left(2\right){\left[\mathit{x}\right]}_0^{\frac{\mathit{\pi }}{4}}-\mathit{I}$$$$2\mathit{I}=\mathit{I}\mathit{n}\left(2\right)[\frac{\mathit{\pi }}{4}-0]$$$$2\mathit{I}=\frac{\mathit{\pi }}{4}\mathit{I}\mathit{n}\left(2\right)$$$$\mathit{I}=\frac{\mathit{\pi }}{8}\mathit{I}\mathit{n}\left(2\right)$$$$$$$$$$$$$$$$$$

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