Re: MX2 2015 Integration Marathon
![](https://latex.codecogs.com/png.latex?\bg_white \text{Let } I = \int_1^3\frac{\ln x}{3+x^2} \text{ d}x. )
Let u = ln x. So x = eu ⇒ dx = eu du.
When x = 1, u = 0, and when x = 3, u = ln 3.
.
Now, using the fact that
, we have
![](https://latex.codecogs.com/png.latex?\bg_white I = \int_0^{\ln 3}\frac{(\ln 3 - u)\text{e}^{\ln 3 - u}}{3+\text{e}^{2{(\ln 3 - u)}}} \text{ d}u. )
Since eln 3 - u = eln 3.e-u = 3e-u, we have
![](https://latex.codecogs.com/png.latex?\bg_white I = \int_0^{\ln 3}\frac{(\ln 3 - u)3\text{e}^{-u}}{3+{(3\text{e}^{-u})}^2} \text{ d}u )
![](https://latex.codecogs.com/png.latex?\bg_white = \int_0^{\ln 3}\frac{(\ln 3) \cdot 3\text{e}^{-u}}{3+9\text{e}^{-2u}} \text{ d}u - \int_0^{\ln 3}\frac{u\cdot 3\text{e}^{-u}}{3+9\text{e}^{-2u}} \text{ d}u )
![](https://latex.codecogs.com/png.latex?\bg_white = \int_0^{\ln 3}\frac{(\ln 3) \text{e}^{-u}}{1+3\text{e}^{-2u}} \text{ d}u - \int_0^{\ln 3}\frac{u\text{e}^{-u}}{1+3\text{e}^{-2u}} \text{ d}u )
(multiplying the numerator and denominator of the integrand of the second integral on the RHS by e2u )
![](https://latex.codecogs.com/png.latex?\bg_white \Rightarrow 2I = \int_0^{\ln 3}\frac{(\ln 3) \text{e}^{-u}}{1+3\text{e}^{-2u}} \text{ d}u. )
This is pretty easy to integrate now. Let v = e-u ⇒ -dv = e-u du and change the limits of integration, so
![](https://latex.codecogs.com/png.latex?\bg_white 2I = - \int_1^{\frac{1}{3}}\frac{\ln 3 }{1+3v^2} \text{ d}v )
![](https://latex.codecogs.com/png.latex?\bg_white \Rightarrow I = \frac{1}{2}\ln 3 \int_\frac{1}{3}^1 \frac{1}{3}\cdot \frac{1}{\frac{1}{3}+v^2} \text{ d}v )
![](https://latex.codecogs.com/png.latex?\bg_white = \frac{1}{6}\ln 3 \times \sqrt{3}\left [ \tan ^{-1}(\sqrt{3}v) \right ]_\frac{1}{3} ^1 )
![](https://latex.codecogs.com/png.latex?\bg_white = \frac{\sqrt{3}\ln 3}{6}\left \{ \tan ^{-1}\sqrt{3}-\tan ^{-1}\frac{\sqrt{3}}{3} \right \} )
![](https://latex.codecogs.com/png.latex?\bg_white = \frac{\sqrt{3}\ln 3}{6}\left \{ \frac{\pi}{3}- \frac{\pi}{6} \right \} )
![](https://latex.codecogs.com/png.latex?\bg_white = \frac{\pi \sqrt{3}\ln 3}{36}. )
lol a 100% magic done by substitution. anyone can find a way to play this magic:
![]()
Let u = ln x. So x = eu ⇒ dx = eu du.
When x = 1, u = 0, and when x = 3, u = ln 3.
Now, using the fact that
Since eln 3 - u = eln 3.e-u = 3e-u, we have
This is pretty easy to integrate now. Let v = e-u ⇒ -dv = e-u du and change the limits of integration, so