Question lives in 4 years

Make a substitution $u=(x-2012)(x-2014)$ , with $du=2(x-2013)dx$ . The integral becomes $\int_{3}^{8}\frac{du}{u}=\ln(\frac{8}{3})$ .

Its a level one problem, who made it level 3 or 2

apply partial fraction

$\frac{x-2013}{(x-2012)(x-2014)}=\frac{1}{2(x-2012)}+\frac{1}{2(x-2014)}$

u will get answer as $ln(\frac{8}{3})$

$(x-2012)(x-2014)=[(x-2013)+1][(x-2013)-1]$ $(x-2012)(x-2014)=(x-2013)^2-1$ Now, let: $u=(x-2013)^2-1, du=2(x-2013)$ Then: $2\int\limits_{2015}^{2016} {\frac{{x - 2013}}{{(x - 2012)(x - 2014)}}dx = } \int\limits_{u(2015)}^{u(2016)} {\frac{{du}}{u}}$ $2\int\limits_{2015}^{2016} {\frac{{x - 2013}}{{(x - 2012)(x - 2014)}}dx = } \left. {\ln u} \right|_{u(2015)}^{u(2016)}$ $2\int\limits_{2015}^{2016} {\frac{{x - 2013}}{{(x - 2012)(x - 2014)}}dx = } \left. {\ln \left[ {\left( {x - 2013} \right)^2 - 1} \right]} \right|_{2015}^{2016}$ $2\int\limits_{2015}^{2016} {\frac{{x - 2013}}{{(x - 2012)(x - 2014)}}dx = } \ln (8/3) \approx 0.9808$

Put x - 2013 = y , then

dx = dy

x - 2012 = y + 1

x - 2014 = y - 1

limits of y :

From 2 to 3

Now , our function to be integrated is

2y/(y + 1)(y - 1) = 1/(y +1) + 1/(y - 1)

The answer is

ln (y + 1) + ln (y - 1) ....................... from 2 to 3

= (ln 4 + ln 2) - (ln 3 + ln 1) = ln (8/3) = 0.9808