# Question and Answers Forum

Integration Questions

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

$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\$$

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

$$\mathrm{1}+{x}^{\mathrm{2}} =\left(\mathrm{1}+{ix}\right)\left(\mathrm{1}−{ix}\right) \\$$

Commented by mohammad17 last updated on 12/Aug/20

$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)=\left({i}+{x}\right)\left(−{i}+{x}\right) \\$$

Commented by mohammad17 last updated on 12/Aug/20

$$\begin{cases}{{u}={ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\Rightarrow{u}^{'} =\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}}\\{{v}^{'} ={dx}\Rightarrow{v}={x}}\end{cases} \\$$$$\\$$$${I}=\left[{xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$$$\\$$$${I}=\left[{xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\left[\mathrm{2}{x}\right]_{\mathrm{0}} ^{\mathrm{1}} +\left[\mathrm{2}{tan}^{−\mathrm{1}} \left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$$\\$$$${I}={ln}\left(\mathrm{2}\right)−\mathrm{2}+\frac{\pi}{\mathrm{2}} \\$$$$\\$$$${by}:\:{mss}:{Mohammad}\:{taha} \\$$

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

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts}\:\mathrm{I}=\left[\mathrm{xln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}.\frac{\mathrm{2x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\$$$$=\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{2}\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\$$$$=\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{2}\:+\mathrm{2}\left[\mathrm{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{2}+\mathrm{2}\left(\frac{\pi}{\mathrm{4}}\right)\:\Rightarrow \\$$$$\mathrm{I}\:=\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{2}\:+\frac{\pi}{\mathrm{2}} \\$$

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

$$\mathrm{I}\:=\:\mathrm{x}.\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\mathrm{1}} \:−\:\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\$$$$\:\:=\:\mathrm{ln}\left(\mathrm{2}\right)\:−\:\mathrm{2}\left(\mathrm{1}\:−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\right) \\$$$$\:\:=\:\mathrm{ln}\left(\mathrm{2}\right)\:+\:\frac{\pi}{\mathrm{2}}\:−\mathrm{2} \\$$$$\bigstar\mathbb{HG}\bigstar \\$$

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

$$\left[{xlog}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\$$$${log}\left(\mathrm{2}\right)−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\$$$${log}\left(\mathrm{2}\right)−\mathrm{2}+\frac{\pi}{\mathrm{2}} \\$$

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

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

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

Brilliant ! Redmiiuser��