8/17/2019 aporte punto 5 y 6
1/2
5.
∫ √ 2+9
2
√ x3
√ x2dx
∫ √ 2+9
3
√ ×3
√ ×2dx
∫√ 2+9 3√ ×.×−2
3dx U =2+9
3
√ ×
∫√ u .du
3 2+9
×1
3
1
3
du
dx=9
3×
−2
3
du
dx=3×
−2
3
du
3=×
−2
3dx
1
3∫ u
1
2 .du=1
3 ( u1
2+1
1
2+1 )+C
1
3 ( u3
2
3
2 )+C =13 (2√ u33 )+C
∫√ 2+9
3
√ ×3
√ ×2
dx=¿
2
9√ (2+9 3√ X )
3
+C
8/17/2019 aporte punto 5 y 6
2/2
6. x∫
x
√ 3− x4dx
∫ ×
√ 3−×4dx
×4
=(×2 )2
U =×2 du
dx=2×
du
2×dx
∫ ×
√ 3−(×2 )2dx
1
2∫
du
√ 3−u2dx
1
2∫
du
√ (√ 3 )2
−u2
a=√ 3
1
2arsen
u
√ 3+C
1
2arsen
×2
√ 3+C