Originally Posted by
polymerase Sure.
Find $\displaystyle \int^4_2 \frac{2x+4}{x^2+4x+5}\,dx$
Now, in order to find this, you must first solve the intergrand. So one way to solve this question is to use the technique of substitution.
So I'm going to let $\displaystyle u=x^2+4x+5$. I have introduced a new variable $\displaystyle u$ in place of $\displaystyle x$. In the orginal function, the limit was x from 2 to 4 but now my intergral is in terms of $\displaystyle u$, which MAY OR MAY NOT go from 2 to 4. Thus to find out what it does go to, you plug in $\displaystyle x=2$ into the equation $\displaystyle u=x^2+4x+5$ to find out what $\displaystyle u$ goes to when $\displaystyle x$ goes to 2. And you do the same for the other limit, 4. In the end, you should find the new limits of u to be from 17 to 37.
Thus, $\displaystyle \int^4_2 \frac{2x+4}{x^2+4x+5}\,dx=\int^{37}_{17} \frac{du}{u}$
You then easily find the answer to be $\displaystyle \ln \left(\frac{37}{17}\right)$ since $\displaystyle \int^{37}_{17} \frac{du}{u}=\ln|u||^{37}_{17}$. You can also after solving the integral, re-substitute then solve again in terms of x from 2 to 4. Same answer!
Hope this helps!