∫logx−11+(log⁡x)22dx=xa(log⁡x)b+t+c

I=∫logx−11+(log⁡x)22dx put logx=t⇒x=et⇒dx=etdt I=∫t-121+t22etdt =∫t2+1-2t1+t22etdt =∫11+t2-2t1+t22etdt =et1+t2+c =x1+logx2+c ⇒t=1,a=1,b=2

∫logx−11+(log⁡x)22dx=xa(log⁡x)b+t+c

I=∫logx−11+(log⁡x)22dx put logx=t⇒x=et⇒dx=etdt I=∫t-121+t22etdt =∫t2+1-2t1+t22etdt =∫11+t2-2t1+t22etdt =et1+t2+c =x1+logx2+c ⇒t=1,a=1,b=2

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$\int {\{\frac{\log x - 1}{1 + (\log x)^{2}}\}}^{2} d x = \frac{x^{a}}{(\log x)^{b} + t} + c$

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a + t = 2

$a^{2} + b^{2} = 5$

a = 2

$a^{2} + t^{2} = 5$

answer is B, D.

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Detailed Solution

$I = \int {\{\frac{\log x - 1}{1 + (\log x)^{2}}\}}^{2} d x p u t \log x = t \Rightarrow x = e^{t} \Rightarrow d x = e^{t} d t I = \int \frac{{(t - 1)}^{2}}{{(1 + t^{2})}^{2}} e^{t} d t = \int \frac{t^{2} + 1 - 2 t}{{(1 + t^{2})}^{2}} e^{t} d t = \int [\frac{1}{(1 + t^{2})} - \frac{2 t}{{(1 + t^{2})}^{2}}] e^{t} d t = \frac{e^{t}}{(1 + t^{2})} + c = \frac{x}{1 + {(\log x)}^{2}} + c \Rightarrow t = 1, a = 1, b = 2$