# Evaluate the integral 2 2 1 5 6 x dx x x x 2 1 x 2 5

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22156xdxxxx2+1x25x6=27x+1+377x627x+1(¿+377x6)dx¿¿¿27ln|x+1|+377ln|x6|+cWA 2, p. 9
24. Evaluate the integral.311dxx1x31=13(x1)x+23(x2+x+1)¿131(x1)x+2x2+x+1dx¿131x1122x+4x2+x+1dx¿131x112(2x+1)x2+x+1123x2+x+1dx3(x+12)2+34dx¿131x1(12)((2x+1)x2+x+1)(12)¿¿13(ln|x1|12ln|x2+x+1|3tan1(2x+13))+cSection 6.68.Determine whether the integral converges or diverges. Find the value of the integral if itconverges.(a) 123xx edxu=3xI=lim∞→11x2e3xdx¿1273u2euduWA 2, p. 10
¿127limR→(u2eu2ueu+2eu)|3R¿527e3127limR→eR(R22R+2)limR→∞eR(R22R+2)¿limR→∞eR(R2+2R+2)¿limR→∞R2+2R+2eR=0I=527e3WA 2, p. 11
(b) 04xxedxu=−4xI=0x e3x4xdx¿1160ueudu¿116limR→∞R0ueudu¿127limR→(ueueu)|0R¿116+116limR→eR(R1)limR→∞eR(R1)¿limR→∞eR(−R1)=0I=11610. Determine whether the integral converges or diverges. Find the value of the integral if itconverges.(a) 0cosx dx¿limR→∞0Rcosx dx¿limR→(sinx)|R0WA 2, p. 12
sinR(¿sin 0)¿limR →∞¿Integral diverges.