Let Î1=∫ limits01 (| ln x|/x2 + 4 x + 1) d x and Î2=∫ limits 1∞( ln x/x2 + 4 x + 1)dx, then

Question

Let Î1=∫ limits01 (| ln x|/x2 + 4 x + 1) d x and Î2=∫ limits 1∞( ln x/x2 + 4 x + 1)dx, then (A) Ι1=Ι2 (B) Ι1>Ι2 (C) Ι1+Ι2=0 (D) Ι1=2Ι2

Tardigrade Admin · Accepted Answer

ln I 2 substitute x=(1/t) ⇒ d x=-(d t/t2) So, I 2=-∫ limits10 ( ln t/t2+4 t+1) ⋅ t2((-d t/t2))=∫ limits01 ((- ln t)/t2+4 t+1) d t= I 1 textas | ln t|=- ln t textin (0,1)

Q. Let $I_{1} = 0 \int 1 \frac{∣ l n x ∣}{x ^{2} + 4 x + 1} d x$ and $I_{2} = 1 \int \infty \frac{l n x}{x ^{2} + 4 x + 1} d x,$ then

$I_{1} = I_{2}$

$I_{1} > I_{2}$

$I_{1} + I_{2} = 0$

$I_{1} = 2 I_{2}$

$ln I_{2}$ substitute $x = \frac{1}{t} \Rightarrow d x = - \frac{d t}{t ^{2}}$
So, $I_{2} = - 1 \int 0 \frac{l n t}{t ^{2} + 4 t + 1} \cdot t^{2} (\frac{- d t}{t ^{2}}) = 0 \int 1 \frac{( - l n t )}{t ^{2} + 4 t + 1} d t = I_{1}$ ${as ∣ ln t ∣ = - ln t in (0, 1)}$

Q. Let I1​=0∫1​x2+4x+1∣lnx∣​dx and I2​=1∫∞​x2+4x+1lnx​dx, then

I1​=I2​

I1​>I2​

I1​+I2​=0

I1​=2I2​