STEP II 1997 question 3 solution

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STEP II 1997 question 3 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1997 question 3 solution

$\displaystyle \frac{ax+b}{x^2+2x+2} + \frac{cx+d}{x^2-2x+2} = \frac{1}{x^4+4}$

Therefore

$\displaystyle (ax+b)(x^2-2x+2) + (cx+d)(x^2+2x+2) = 1$

Expand out

$\displaystyle ax^3 -2ax^2 + 2ax + bx^2 -2bx + 2b + cx^3 +2cx^2 + 2cx + dx^2 + 2dx + 2d = 1$

$\displaystyle (a+c)x^3 (-2a+b+2c+d)x^2 + (2a-2b+2c+2d)x + 2b + 2d = 1$

Comparing coefficients

a + c = 0 - 1

-2a+b+2c+d = 0 - 2

2a-2b+2c+2d = 0 - 3

2b+2d = 1 - 4

a = -c - Substituting this into 2 and 3

b + 4c + d = 0

b = d - Substituting this into 4

4b = 1

$b = \frac{1}{4}$

4d = 1

$d = \frac{1}{4}$

Substituting values of b and d into 2

$\frac{1}{2} + 4c = 0$

$c = -\frac{1}{8}$ - Substituting back into 1

$a = \frac{1}{8}$

$b = \frac{1}{4}$

$c = -\frac{1}{8}$

$d = \frac{1}{4}$

Therefore

$\displaystyle \frac{1}{x^4+4} = \frac{\frac{1}{8}x+\frac{1}{4}}{x^2+2x+2} + \frac{\frac{1}{4}-\frac{1}{8}x}{x^2-2x+2}$

$\displaystyle = \frac{x+2}{8x^2+16x+16} + \frac{2-x}{8x^2-16x+16}$

$\displaystyle \int \frac{1}{x^4+4} = \int \frac{x+2}{8x^2+16x+16} + \frac{2-x}{8x^2-16x+16}$

$\displaystyle = \frac{1}{8}\int \frac{x+2}{x^2+2x+2} + \frac{2-x}{x^2-2x+2}$

Consider $\displaystyle \int \frac{x+2}{x^2+2x+2}$

$\displaystyle \int \frac{x+2}{x^2+2x+2} = \int \frac{2x+2}{x^2+2x+2} - \frac{x}{x^2+2x+2}$

$\displaystyle = \ln(x^2+2x+2) - \int \frac{x}{x^2+2x+2}$

Consider $\displaystyle \int \frac{x}{x^2+2x+2}$

$\displaystyle \int \frac{x}{x^2+2x+2} = \int \frac{2x+2}{x^2+2x+2} - \frac{2}{x^2+2x+2} - \frac{x}{x^2+2x+2}$

$\displaystyle 2\int \frac{x}{x^2+2x+2} = ln (x^2+2x+2) - \int \frac{2}{x^2+2x+2}$

$\displaystyle = ln (x^2+2x+2) - \int \frac{2}{(x+1)^2+1}$

let u = x+1

$\displaystyle \frac{du}{dx} = 1$

du = dx

$\displaystyle = \ln (x^2+2x+2) - 2\int\frac{1}{u^2+1}$

$\displaystyle = \ln(x^2+2x+2) - 2\arctanu$

$\displaystyle = \ln(x^2+2x+2) - 2\arctan(x+1)$

$\displaystyle 2\int \frac{x}{x^2+2x+2} = \ln(x^2+2x+2) - 2\arctan(x+1)$

$\displaystyle \int \frac{x}{x^2+2x+2} = \frac{1}{2}\ln(x^2+2x+2) - \arctan(x+1)$

Therefore

$\displaystyle \int \frac{x+2}{x^2+2x+2} = \ln(x^2+2x+2) - \left[\frac{1}{2}ln(x^2+2x+2) - arctan(x+1)\right]$

$= \frac{1}{2}ln(x^2+2x+2) + \arctan(x+1)$

Consider $\displaystyle \int \frac{2-x}{x^2-2x+2}$

$\displaystyle \int \frac{2-x}{x^2-2x+2} = \int \frac{2x-2}{x^2-2x+2} +\int \frac{4}{x^2-2x+2} - \int \frac{3x}{x^2-2x+2}$

$\displaystyle = \ln (x^2-2x+2) + 4\int \frac{1}{(x-1)^2+1} - 3 \int \frac{x}{x^2-2x+2}$

Consider $\displaystyle \int \frac{1}{(x-1)^2+1}$

u = x-1

du = dx

$\displaystyle \int \frac{1}{(x-1)^2+1} = \int \frac{1}{u^2+1}$

$= \arctan(x-1)$

$\displaystyle \int \frac{2-x}{x^2-2x+2} = \ln(x^2-2x+2) +4\arctan(x-1) - \int \frac{3x}{x^2-2x+2}$

Consider $\displaystyle \int \frac{3x}{x^2-2x+2}$

$\displaystyle \int \frac{3x}{x^2-2x+2} = \int \frac{2x-2}{x^2-2x+2} + \int \frac{x}{x^2-2x+2} + \int \frac{2}{x^2-2x+1}$

$2\displaystyle \int \frac{x}{x^2-2x+2} = \ln (x^2-2x+2) + 2\arctan(x-1)$

$\displaystyle \int \frac{x}{x^2-2x+2} = \frac{1}{2}\ln (x^2-2x+2) + \arctan(x-1)$

Therefore

$\displaystyle \int \frac{2-x}{x^2-2x+2} = \ln(x^2-2x+2) +4\arctan(x-1) -3\left[\frac{1}{2}ln (x^2-2x+2) + arctan(x-1)\right]$

Finally

$\displaystyle \int \frac{1}{x^4+4} = \frac{1}{8} \int \frac{x+2}{x^2+2x+2} + \frac{2-x}{x^2-2x+2}$

[Unparseable or potentially dangerous latex formula. Error 5 : 1229x52]

$\displaystyle = \left[\frac{1}{8}\ln5 - \frac{1}{8}\left(\frac{1}{2}\ln5 - \arctan2\right)+ \frac{1}{8}\ln1 + \frac{1}{2}\ln0 - \frac{3}{8}\left(\frac{1}{2}\ln1+\arctan0\right)\right] - \left[\frac{1}{8}\ln2 - \frac{1}{8}\left(\frac{1}{2}\ln2 - \arctan(1)\right)+\frac{1}{8}\ln2 + \frac{1}{2}\arctan(-1) -\frac{3}{8}\left(\frac{1}{2}\ln2+\arctan(-1)\right)\right]$

$\displaystyle = \left[\frac{1}{16}\ln5 +\frac{1}{8}\arctan(2)\right] - \left[\frac{1}{16}\ln2 + \frac{\pi}{32} + \frac{1}{8}\ln2 -\frac{\pi}{8} - \frac{3}{16}\ln2 +\frac{3\pi}{32}\right]$

$\displaystyle = \frac{1}{16}\ln5 + \frac{1}{8}\arctan2$

Solution by insparato

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