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STEP II 1997 question 3 solutionTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1997 question 3 solution
Therefore
Expand out
Comparing coefficients
Substituting values of b and d into 2
So
Therefore
Consider
Consider
let u = x+1
Therefore
Consider
Consider
Consider
Therefore
Finally
Unparseable or potentially dangerous latex formula. Error 5: Image dimensions are out of bounds: 1236x53
\displaystyle = \left[\frac{1}{8}\ln(x^2+2x+2) -\frac{1}{8}\left(\frac{1}{2}ln(x^2+2x+2)-arctan(x+1)\right) + \frac{1}{8}\ln(x^2-2x+2) + \frac{1}{2}\arctan(x-1) - \frac{3}{8}\left(\frac{1}{2}\ln(x^2-2x+2) + arctan(x-1)\right)\right]_0^1 Unparseable or potentially dangerous latex formula. Error 5: Image dimensions are out of bounds: 1396x48
\displaystyle = \left[\frac{1}{8}\ln5 - \frac{1}{8}\left(\frac{1}{2}\ln5 - \arctan2\right)+ \frac{1}{8}\ln1 + \frac{1}{2}\ln0 - \frac{3}{8}\left(\frac{1}{2}\ln1+\arctan0\right)\right] - \left[\frac{1}{8}\ln2 - \frac{1}{8}\left(\frac{1}{2}\ln2 - \arctan(1)\right)+\frac{1}{8}\ln2 + \frac{1}{2}\arctan(-1) -\frac{3}{8}\left(\frac{1}{2}\ln2+\arctan(-1)\right)\right]
Solution by insparato |
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![\displaystyle \int \frac{x+2}{x^2+2x+2} = \ln(x^2+2x+2) - \left[\frac{1}{2}ln(x^2+2x+2) - arctan(x+1)\right]](http://www.thestudentroom.co.uk/latexrender/pictures/67/67b01e9df169dac324683cb8cf773b5e.png "\displaystyle \int \frac{x+2}{x^2+2x+2} = \ln(x^2+2x+2) - \left[\frac{1}{2}ln(x^2+2x+2) - arctan(x+1)\right]")
![\displaystyle \int \frac{2-x}{x^2-2x+2} = \ln(x^2-2x+2) +4\arctan(x-1) -3\left[\frac{1}{2}ln (x^2-2x+2) + arctan(x-1)\right]](http://www.thestudentroom.co.uk/latexrender/pictures/ac/ac0b8102425251e283dc94f8c3ef9669.png "\displaystyle \int \frac{2-x}{x^2-2x+2} = \ln(x^2-2x+2) +4\arctan(x-1) -3\left[\frac{1}{2}ln (x^2-2x+2) + arctan(x-1)\right]")
![\displaystyle = \left[\frac{1}{16}\ln5 +\frac{1}{8}\arctan(2)\right] - \left[\frac{1}{16}\ln2 + \frac{\pi}{32} + \frac{1}{8}\ln2 -\frac{\pi}{8} - \frac{3}{16}\ln2 +\frac{3\pi}{32}\right]](http://www.thestudentroom.co.uk/latexrender/pictures/6f/6f83acafa1043ef618dbedb503da9c68.png "\displaystyle = \left[\frac{1}{16}\ln5 +\frac{1}{8}\arctan(2)\right] - \left[\frac{1}{16}\ln2 + \frac{\pi}{32} + \frac{1}{8}\ln2 -\frac{\pi}{8} - \frac{3}{16}\ln2 +\frac{3\pi}{32}\right]")
