31B(F07)_Midterm2_Solutions - MATH 31B Lecture 4 Fall 2007 Solutions to Midterm 2 This is a closed-book and closed-note examination Calculators are not

# 31B(F07)_Midterm2_Solutions - MATH 31B Lecture 4 Fall 2007...

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MATH 31B - Lecture 4 - Fall 2007Solutions to Midterm 2 - November 16, 2007This is a closed-book and closed-note examination.Calculators are not allowed.Please show all your work.Use only the paper provided. You may write on the back if you need morespace, but clearly indicate this on the front.There are 6 problems for a total of 100 points.POINTS:1.2.3.4.5.6.TOTAL:1
21. (15 points)Evaluate the integralZ3x+ 1x4+x3+x2dxSolution.Here we do a partial fraction decomposition and write:3x+ 1x4+x3+x2=3x+ 1x2(x2+x+ 1)=Ax+Bx2+Cx+Dx2+x+ 1Comparing the numerators of both sides gives us:Ax(x2+x+ 1) +B(x2+x+ 1) + (Cx+D)(x2) = 3x+ 1Comparing the coefficients forx3,x2,x, and the constants gives us thefollowing equations:A+C= 0A+B+D= 0A+B= 3B= 1Solving the equations gives usA= 2,B= 1,C=-2,D=-3 Now we cantry to solve our integral.Z3x+ 1x4+x3+x2dx=Z2xdx+Z1x2dx+Z-2-3xx2+x+ 1dx= 2 ln|x| -1/x+Z-2x-3x2+x+ 1dxLet us complete the square to evaluate the last integral. We letu= (x+1/2)thenZ-2x-3x2+x+ 1=Z-2u-2u2+ 3/4du=Z-2uu2+ 3/4+Z-2u2+ 3/4du=-ln|u2+ 3/4| -433tan-1(u/p3/4).