Question Number 136644 by mnjuly1970 last updated on 24/Mar/21
![𝛗=∫_0 ^( 1) ((ln(x^2 +1))/x^2 )dx f(a)=∫_0 ^( 1) ((log(ax^2 +1))/x^2 )dx f ′(a)=∫_0 ^( 1) (x^2 /((ax^2 +1)x^2 ))dx=(1/a)∫_0 ^( 1) (dx/(x^2 +((1/( (√a))))^2 )) =((√a)/a)[tan^(−1) (x(√a) )]_0 ^1 =((√a)/a)tan^(−1) ((√a)) f(a)=^((√a) =u) ∫2tan^(−1) (u)du =2{u.tan^(−1) (u)−∫(u/(1+u^2 ))du}+C =2(√a) tan^(−1) ((√a) )−ln(1+a)+C f(0)=0=0+C⇒C=0 f(1)=𝛗=2((π/4))−ln(2)=(π/2)−ln(2)](https://www.tinkutara.com/question/Q136644.png)
$$ \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{log}\left({ax}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:{f}\:'\left({a}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\left({ax}^{\mathrm{2}} +\mathrm{1}\right){x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{dx}}{{x}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{{a}}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\sqrt{{a}}}{{a}}\left[{tan}^{−\mathrm{1}} \left({x}\sqrt{{a}}\:\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\sqrt{{a}}}{{a}}{tan}^{−\mathrm{1}} \left(\sqrt{{a}}\right) \\ $$$$\:\:{f}\left({a}\right)\overset{\sqrt{{a}}\:={u}} {=}\int\mathrm{2}{tan}^{−\mathrm{1}} \left({u}\right){du} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left\{{u}.{tan}^{−\mathrm{1}} \:\left({u}\right)−\int\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\right\}+{C} \\ $$$$\:=\mathrm{2}\sqrt{{a}}\:{tan}^{−\mathrm{1}} \left(\sqrt{{a}}\:\right)−{ln}\left(\mathrm{1}+{a}\right)+{C} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}=\mathrm{0}+{C}\Rightarrow{C}=\mathrm{0} \\ $$$$\:\:\:{f}\left(\mathrm{1}\right)=\boldsymbol{\phi}=\mathrm{2}\left(\frac{\pi}{\mathrm{4}}\right)−{ln}\left(\mathrm{2}\right)=\frac{\pi}{\mathrm{2}}−{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$
Answered by mnjuly1970 last updated on 24/Mar/21
![method 2: 𝛗=^(i.b.p) [((−1)/x)ln(x^2 +1)]_0 ^1 +∫_0 ^1 ((2x)/(x(1+x^2 )))dx =−log(2)+(π/2) ..](https://www.tinkutara.com/question/Q136646.png)
$${method}\:\mathrm{2}: \\ $$$$\:\boldsymbol{\phi}\overset{{i}.{b}.{p}} {=}\left[\frac{−\mathrm{1}}{{x}}{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=−{log}\left(\mathrm{2}\right)+\frac{\pi}{\mathrm{2}}\:\:.. \\ $$