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Question Number 107790 by Ar Brandon last updated on 12/Aug/20

$${\int }_0^1\ln (1+{\mathrm{x}}^2)\mathrm{dx}$$

Commented by prakash jain last updated on 12/Aug/20

$$1+x^2=(1+ix)(1-ix)$$

Commented by mohammad17 last updated on 12/Aug/20

$$(1+x^2)=(i+x)(-i+x)$$

Commented by mohammad17 last updated on 12/Aug/20

$$\{\begin{array}{l}u=ln(1+x^2)\Rightarrow u^{{}^{\prime }}=\frac{2x}{1+x^2}dx\\ v^{{}^{\prime }}=dx\Rightarrow v=x\end{array}$$$$$$$$I={\left[xln(1+x^2)\right]}_0^1-2{\int }_0^1\frac{x^2+1-1}{1+x^2}dx$$$$$$$$I={\left[xln(1+x^2)\right]}_0^1-{\left[2x\right]}_0^1+{\left[2{tan}^{-1}\left(x\right)\right]}_0^1$$$$$$$$I=ln\left(2\right)-2+\frac{\pi }{2}$$$$$$$$by:mss:Mohammadtaha$$

Commented by Ar Brandon last updated on 12/Aug/20

Thanks Sir

Commented by Ar Brandon last updated on 12/Aug/20

Thank you

Answered by mathmax by abdo last updated on 12/Aug/20

$$\mathrm{I}={\int }_0^1\ln (1+{\mathrm{x}}^2)\mathrm{dx}\mathrm{by}\mathrm{parts}\mathrm{I}={\left[\mathrm{xln}(1+{\mathrm{x}}^2)\right]}_0^1-{\int }_0^1\mathrm{x}.\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=\ln \left(2\right)-2{\int }_0^1\frac{1+{\mathrm{x}}^2-1}{1+{\mathrm{x}}^2}\mathrm{dx}=\ln \left(2\right)-2+2{\int }_0^1\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}$$$$=\ln \left(2\right)-2+2{\left[\mathrm{arctanx}\right]}_0^1=\ln \left(2\right)-2+2\left(\frac{\pi }{4}\right)\Rightarrow$$$$\mathrm{I}=\ln \left(2\right)-2+\frac{\pi }{2}$$

Commented by Ar Brandon last updated on 12/Aug/20

Merci monsieur��

Commented by mathmax by abdo last updated on 13/Aug/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Answered by hgrocks last updated on 12/Aug/20

$$\mathrm{I}=\mathrm{x}.\ln (1+{\mathrm{x}}^2){\mid }_0^1-2\underset{0}{\stackrel{1}{\int }}\frac{{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=\ln \left(2\right)-2(1-\underset{0}{\stackrel{1}{\int }}\frac{1}{1+{\mathrm{x}}^2}\mathrm{dx})$$$$=\ln \left(2\right)+\frac{\pi }{2}-2$$$$★\mathbb{HG}★$$

Answered by Dwaipayan Shikari last updated on 12/Aug/20

$${\left[xlog(1+x^2)\right]}_0^1-2{\int }_0^1\frac{x^2}{1+x^2}$$$$log\left(2\right)-2{\int }_0^{1}1-\frac{1}{1+x^2}$$$$log\left(2\right)-2+\frac{\pi }{2}$$

Commented by Ar Brandon last updated on 12/Aug/20

Hi ���� Thanks

Commented by Dwaipayan Shikari last updated on 12/Aug/20

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Answered by hgrocks last updated on 12/Aug/20

$$\mathrm{Method}2:\mathrm{Using}\mathrm{Series}$$$$$$$$\mathrm{I}=\underset{0}{\stackrel{1}{\int }}\underset{\mathrm{r}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{r}-1}}{\mathrm{r}}{\mathrm{x}}^{2\mathrm{r}}\mathrm{dx}$$$$=\underset{\mathrm{r}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{r}-1}}{(2\mathrm{r}+1)\mathrm{r}}$$$$=2\underset{\mathrm{r}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{r}-1}}{(2\mathrm{r}+1)\left(2\mathrm{r}\right)}$$$$=2\underset{\mathrm{r}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{r}-1}}{2\mathrm{r}}-2\underset{\mathrm{r}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{r}-1}}{(2\mathrm{r}+1)}$$$${\mathrm{S}}_1=(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......)=\ln \left(2\right)$$$${\mathrm{S}}_2=(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}...........)=1-$$$${\tan }^{-1}\left(1\right)$$$$$$$$\mathrm{So}\mathrm{I}={\mathrm{S}}_1-{2\mathrm{S}}_2=\ln \left(2\right)+\frac{\pi }{2}-2$$

Commented by Ar Brandon last updated on 12/Aug/20

Brilliant ! Redmiiuser��

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