MATH 115 Midterm: MATH 115 UPenn 115Fall 07 Exams
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4 x2 dx x = 2 sin dx = 2 cos d . 4 x2 = 4 4 sin2 = 4 cos2 . Then the integral we have to solve is. 4 x2 dx = z (2 sin + 1)2 cos d . = z (2 sin + 1) d . Now we have to plug back in for . From the equations above, or using the triangle, we have cos = 4 x2, and of course = arcsin(cid:0) x. 4 x2 dx = 2 4 x2 + arcsin(cid:16) x. Note that x2 + 4x + 3 = (x + 1)(x + 3), so using partial fractions, we get. Solving for a and b in the equation. 4 = a(x + 3) + b(x + 1) we get a = 2, b = 2. 4 x2 + 4x + 3 dx = z (cid:18) 2. 2 x + 3(cid:19) dx x + 1 dx x + 1 2z dx.
