The idealized integral

Calculus Level 5

$\large \int_1^\infty \frac{2x^3-1}{x^6 + 2x^3 + 9x^2 + 1} \, dx$

If the integral above equals to $\frac AB \cot^{-1} \left( \frac CD \right)$ for positive integers $A,B,C,D$ such that $\gcd(A,B) = \gcd(C,D) = 1$ , evaluate $2(A+B+C+D)$ .

The answer is 18.

1 solution

Vishnu C
Jul 1, 2015

Given integral = $\large\displaystyle\int _{1} ^{\infty} {\frac {2x^3-1}{(x^3+1)^2+9x^2}} dx$

$\large\displaystyle\Rightarrow \int _{1} ^{\infty} {\frac {2x^3-1}{x^2((x^2+\frac 1 x )^2+9)}} dx$

$\large\displaystyle\Rightarrow \int _{1} ^{\infty} {\frac {2x-\frac 1 {x^2}}{(x^2+\frac 1 x )^2+9}} dx$

Substituting $x^2+\frac 1 x = t \Rightarrow dx = \frac {dt} {(2x-\frac 1 {x^2})}$ we get,

$\frac {1} { 3} \tan ^{-1} (\frac {x^2 + \frac 1 x } 3 )|^{\infty} _{1}\\ \Rightarrow \frac 1 3 ( \frac {\pi} 2 -\tan ^{-1} (\frac 2 3))\\ \Rightarrow \frac 1 3 \cot ^{-1} (\frac 23 ) \quad (\because \tan^{-1}(x)+\cot ^{-1} (x) = \frac {\pi} 2)\\ \Rightarrow A =1; B=3; C =2; D =3.\\ \Rightarrow 2(A+B+C+D) = \boxed {18}$

Moderator note:

Careful there, you need to write those $dx$ 's. Nice work though!

A very interesting definite integral. Congratulation!

Lu Chee Ket - 5 years, 7 months ago

The idealized integral

The answer is 18.

1 solution

Moderator note:

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