Valentine's Day! 2

Calculus Level 3

$\large \int_2^{14} \dfrac{1402}{x^2+16x+28} \, dx$

If the integral above is equal to $\dfrac ab \ln\left( \dfrac cd\right)$ , where $a,b,c$ and $d$ are positive integers and $\gcd(a,b) = \gcd(c,d) = 1$ , find $a+b+c+d$ .

The answer is 730.

1 solution

Rishabh Jain
Feb 12, 2016

$x^2+16x+28=(x+14)(x+2)$ $\large\therefore \large \int _{ 2 }^{ 14 } \dfrac{ 1402 ~dx}{ x^ 2 +16x+28 }$ $\large=\dfrac{1402}{12}\int_{2}^{14}(\dfrac{1}{x+2}-\dfrac{1}{x+14})dx$ $\large =\dfrac{701}{6}\ln (\dfrac{x+2}{x+14})\huge|_{\small{2}}^{\small{14}}$ $\Large= \dfrac{701}{6}\ln (\dfrac{16}{7})$ $\Large 701+6+16+7=\huge\boxed{\color{#007fff}{730}}$

Valentine's Day! 2

The answer is 730.

1 solution

0 pending reports