Let Ι_1=∫01(ln⁡x)/(x^2+4x+1)dx and Ι_2=∫1∞(ln⁡x)/(x^2+4x+1)dx, then

Correct answer: (A) Ι_1=Ι_2. In Ι_2 substitute x=1/t⇒dx=-dt/(t^2) So, Ι_2=-∫10(ln⁡t)/(t^2+4t+1)⋅t^2(-dt)/(t^2)=∫01(-ln⁡t)/(t^2+4t+1)dt=Ι_1 as ln⁡t=-ln⁡t in 0,1

Q19 / 116 · NTA Abhyas (JEE Main) 2020 · Mathematics · Definite Integration

Let $Ι_{1} = \int_{0}^{1} \frac{|\ln x|}{x^{2} + 4 x + 1} d x$ and $Ι_{2} = \int_{1}^{\infty} \frac{\ln x}{x^{2} + 4 x + 1} d x,$ then

Correct answer
A
$Ι_{1} = Ι_{2}$
B
$Ι_{1} > Ι_{2}$
C
$Ι_{1} + Ι_{2} = 0$
D
$Ι_{1} = 2 Ι_{2}$

Solution

Ι_{2}

substitute

x = \frac{1}{t} \Rightarrow d x = - \frac{d t}{t^{2}}

So,

Ι_{2} = - \int_{1}^{0} \frac{\ln t}{t^{2} + 4 t + 1} \cdot t^{2} (\frac{- d t}{t^{2}}) = \int_{0}^{1} \frac{(- \ln t)}{t^{2} + 4 t + 1} d t = Ι_{1}

\{a s |\ln t| = - \ln t i n (0,1)\}

Hence, the correct answer is option: 'A'.

Let Ι1=∫01ln⁡xx2+4x+1dx and Ι2=∫1∞ln⁡xx2+4x+1dx, then

Solution

More Definite Integration Questions

Let $Ι_{1} = \int_{0}^{1} \frac{|\ln x|}{x^{2} + 4 x + 1} d x$ and $Ι_{2} = \int_{1}^{\infty} \frac{\ln x}{x^{2} + 4 x + 1} d x,$ then