Answer and Explain Please.

MTH 232 Practice questions for MIDTERM FALL 21 INTEGRATION BY PART: la./ / x cos(2x) dx = 6./ / x3 In x dx c./ (3x+9)et dr = 2a./ tan 1(x) dx = b./ ( e" sin(z) dx = c./ Inde = 3a./ Inxdx = b./ [ ze' dx = c/ f xledr = 4a./ / cos-1(x) dx = b./ [ xe ' dx = TRIGONOMETRIC INTEGRALS: la./a./ sin" x da = b./ / cosex dx = c./ / sin? x cos? x 2a./ sin' x dx = b./ / sinx cos' r dx = c./ sin' r cos' x 3a./ sin' cos3 x dx = b./ / sin r cost x dx = c./ tanx dr = Trigonometric substitutions : Val - x2 : x = a sin 0, 1va2 + x2 : r = atan 0, Vx2 - a2 : T = asec 0 a./ VI - dx = b.../ J 1+72 de = 2.a./ dx = dx = 6./ xvx2 + 16 In question 3 use trigonometric substitution. Which of the two has an easier solution with substitution u = 1 - x2 ? 3.a./ = dx = 6./ / x? VI - x' dr =Use the method of partial fractions in the following problems: 2x - 1 4.a./ x(2x + 1) 6./ - dix = x2 - 5x + 6 dx 6./ / x + 3 5.a./ = dx = (x - 2) (x + 2) (x - 3) x + 4 dx x2 - x+1 2 6.a./ 2 2 - 25 dx = x(x + 1) c./ de = x(x2 + 1 ) Use any methods, you might need more than one to solve a problem. dx 7.a./ b./ / Vx dx = x2 + 2.x + 2 x+ 3 8.a./ In (x2 + 4) dx = 6./ Vx - 1 dx = 9 . dx = Vx+ 2 10.a./ (3sec(x)-cos(x)) dx = b./ sin (x) tan(x) dx =