= 4x^2-2x+1 > 0 \quad \forall x\in[0,1] $

For positive arguments,

$ \cot^{-1}(u)=\tan^{-1}\!\left(\frac{1}{u}\right) $

$ \int_0^1 4\cot^{-1}(4x^2-2x+1)\,dx =4\int_0^1 \tan^{-1}\!\left(\frac{1}{4x^2-2x+1}\right)dx $

Using the identity $\tan^{-1}(1/u)=\frac{\pi}{2}-\tan^{-1}(u)$ (for $u>0$):

$ =4\left[\frac{\pi}{2}-\int_0^1 \tan^{-1}(4x^2-2x+1)\,dx\right] $

Using integration by parts and standard integrals of rational functions, one obtains:

$ \int_0^1 \tan^{-1}(4x^2-2x+1)\,dx = \tan^{-1}(3)-\frac{1}{4}\ln 5 $

Substitute back:

$ \begin{aligned} I &=4\left[\frac{\pi}{2}-\tan^{-1}(3)+\frac{1}{4}\ln 5\right] \\ &=4\tan^{-1}\!\left(\frac{1}{3}\right)+\ln 5 \end{aligned} $

Using the identity:

$ \tan^{-1}(2)=2\tan^{-1}\!\left(\frac{1}{3}\right) $

we get:

$ I = 4\tan^{-1}(2)-\ln 5 $

Comparing with

$ a\tan^{-1}(2)-b\ln(5) $

we have:

$ a=4,\quad b=1 $

$ 2a+b = 2(4)+1 = \boxed{9} $

19:T7c2,

हमें दिया गया है

$ \int_0^1 4\cot^{-1}(1-2x+4x^2)\,dx = a\tan^{-1}(2)-b\ln(5), \qquad a,b\in\mathbb N $

$ 1-2x+4x^2 = 4x^2-2x+1 > 0 \quad \forall x\in[0,1] $

धनात्मक मानों के लिए,

$ \cot^{-1}(u)=\tan^{-1}\!\left(\frac{1}{u}\right) $

इसलिए

$ \int_0^1 4\cot^{-1}(4x^2-2x+1)\,dx =4\int_0^1 \tan^{-1}\!\left(\frac{1}{4x^2-2x+1}\right)dx $

सर्वसमिका $\tan^{-1}(1/u)=\frac{\pi}{2}-\tan^{-1}(u)$ ($u>0$ के लिए) का उपयोग करने पर:

$ =4\left[\frac{\pi}{2}-\int_0^1 \tan^{-1}(4x^2-2x+1)\,dx\right] $

खंडशः समाकलन

Definite Integration

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