Question

Let $Ι_{1}=\int \limits_{0}^{1} \frac{\left|\ln x\right|}{x^{2} + 4 x + 1} d x$ and $Ι_{2}=\int\limits _{1}^{\infty}\frac{\ln x}{x^{2} + 4 x + 1}dx,$ then

• A. $Ι_{1}=Ι_{2}$
• B. $Ι_{1}>Ι_{2}$
• C. $Ι_{1}+Ι_{2}=0$
• D. $Ι_{1}=2Ι_{2}$

Answer 1

Answer: (A)

$\ln I _{2}$ substitute $x=\frac{1}{t} \Rightarrow d x=-\frac{d t}{t^{2}}$
So, $I _{2}=-\int\limits_{1}^{0} \frac{\ln t}{t^{2}+4 t+1} \cdot t^{2}\left(\frac{-d t}{t^{2}}\right)=\int\limits_{0}^{1} \frac{(-\ln t)}{t^{2}+4 t+1} d t= I _{1}$ $\{\text{as} |\ln t|=-\ln t \text{in} (0,1)\}$