I'm bored (Part 2)

Calculus Level 5

The integral $\int _{ 1}^{ 5 } { \frac { dx }{ x^4 + 4x^3 + 9x^2 + 10x } }$ can be expressed as $\frac { \alpha }{ \beta } \pi -\frac { \gamma }\delta { \arctan { \varepsilon } }+\frac { \zeta }{ \eta } \ln { \frac { \theta }{ \iota } }$ where $\alpha$ , $\beta$ , $\gamma$ , $\delta$ , $\varepsilon$ , $\zeta$ , $\eta$ , $\theta$ , and $\iota$ are all positive integers. Also $gcd(\alpha, \beta) = gcd(\gamma, \delta) = gcd(\zeta, \eta) = gcd(\theta, \iota) = 1$ .

Find $\alpha +\beta +\gamma +\delta +\varepsilon +\zeta +\eta +\theta +\iota$

The answer is 88.

1 solution

Tom Engelsman
Jun 3, 2018

The above integrand decomposes into:

$\frac{1}{x^4 + 4x^3 + 9x^2 + 10x} = \frac{1}{10x} - \frac{1}{10(x+2)} - \frac{1}{5(x^2 + 2x + 5)}$ ;

which upon integration gives:

$\int_{1}^{5} \frac{1}{10x} - \frac{1}{10(x+2)} - \frac{1}{5(x^2 + 2x + 5)} dx = \int_{1}^{5} \frac{1}{10x} - \frac{1}{10(x+2)} - \frac{1}{5((x + 1)^{2} + 2^{2})} dx$ ;

or $\frac{1}{10} \cdot ln(\frac{x}{x+2}) - \frac{1}{10} \cdot arctan(\frac{x+1}{2})|_{1}^{5} = \frac{1}{10}[ln(\frac{5}{7}) - ln(\frac{1}{3})] - \frac{1}{10}[arctan(3) - arctan(1)] = \boxed{\frac{1}{10} \cdot ln(\frac{15}{7}) - \frac{1}{10} \cdot arctan(3) + \frac{\pi}{40}}.$

I'm bored (Part 2)

The answer is 88.

1 solution

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