Question Number 30774 by abdo imad last updated on 25/Feb/18
$${find}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}+{tx}^{\mathrm{2}} \right){dx}\:\:{with}\:{t}>\mathrm{0} \\ $$
Commented by abdo imad last updated on 26/Feb/18
![let put (√t) x=u ⇒f(t)= ∫_0 ^(√t) ln(1+u^2 )(du/( (√t))) =(1/( (√t))) ∫_0 ^(√t) ln(1+u^2 )du but we have by parts ∫_0 ^(√t) ln(1+u^2 )du= [uln(1+u^2 )]_0 ^(√t) −∫_0 ^(√t) u((2u)/(1+u^2 ))du =(√t) ln(1+t) −2∫_0 ^(√t) ((1+u^2 −1)/(1+u^2 ))du =(√t) ln(1+t) −2(√t) +2 ∫_0 ^(√t) (du/(1+u^2 )) =(√t) ln(1+t)−2(√t) +2 arctan((√t))⇒ f(t)= ln(1+t) +((2 arctan((√t)))/( (√t))) −2 .](https://www.tinkutara.com/question/Q30846.png)
$${let}\:{put}\:\sqrt{{t}}\:{x}={u}\:\Rightarrow{f}\left({t}\right)=\:\int_{\mathrm{0}} ^{\sqrt{{t}}} \:{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\frac{{du}}{\:\sqrt{{t}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{t}}}\:\int_{\mathrm{0}} ^{\sqrt{{t}}} {ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right){du}\:\:{but}\:{we}\:{have}\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\sqrt{{t}}} {ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right){du}=\:\left[{uln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\sqrt{{t}}} \:−\int_{\mathrm{0}} ^{\sqrt{{t}}} {u}\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$=\sqrt{{t}}\:{ln}\left(\mathrm{1}+{t}\right)\:−\mathrm{2}\int_{\mathrm{0}} ^{\sqrt{{t}}} \frac{\mathrm{1}+{u}^{\mathrm{2}} \:−\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$=\sqrt{{t}}\:{ln}\left(\mathrm{1}+{t}\right)\:−\mathrm{2}\sqrt{{t}}\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\sqrt{{t}}} \:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\sqrt{{t}}\:{ln}\left(\mathrm{1}+{t}\right)−\mathrm{2}\sqrt{{t}}\:\:+\mathrm{2}\:{arctan}\left(\sqrt{{t}}\right)\Rightarrow \\ $$$${f}\left({t}\right)=\:{ln}\left(\mathrm{1}+{t}\right)\:+\frac{\mathrm{2}\:{arctan}\left(\sqrt{{t}}\right)}{\:\sqrt{{t}}}\:−\mathrm{2}\:. \\ $$
Answered by sma3l2996 last updated on 25/Feb/18
![let u=tx f(t)=∫_0 ^t ln(1+u^2 )du =[uln(1+u^2 )]_0 ^t −2∫_0 ^t (u^2 /(1+u^2 ))du =tln(1+t^2 )−2∫_0 ^t (1−(1/(1+u^2 )))du =tln(1+t^2 )−2[u−tan^(−1) (u)]_0 ^t f(t)=tln(1+t^2 )−2t+2tan^(−1) (t)](https://www.tinkutara.com/question/Q30788.png)
$${let}\:\:{u}={tx} \\ $$$${f}\left({t}\right)=\int_{\mathrm{0}} ^{{t}} {ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right){du} \\ $$$$=\left[{uln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{{t}} −\mathrm{2}\int_{\mathrm{0}} ^{{t}} \frac{{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$={tln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)−\mathrm{2}\int_{\mathrm{0}} ^{{t}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\right){du} \\ $$$$={tln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)−\mathrm{2}\left[{u}−{tan}^{−\mathrm{1}} \left({u}\right)\right]_{\mathrm{0}} ^{{t}} \\ $$$${f}\left({t}\right)={tln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)−\mathrm{2}{t}+\mathrm{2}{tan}^{−\mathrm{1}} \left({t}\right) \\ $$
Commented by abdo imad last updated on 26/Feb/18
$${if}\:{u}={tx}\:{we}\:{have}\:{ln}\left(\mathrm{1}+{tx}^{\mathrm{2}} \right)\neq{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\:{and}\:{your}\:{answer} \\ $$$${sir}\:{sma}\mathrm{3}\:{is}\:{not}\:{corect}… \\ $$