5.4 - d dx x 2 +1 + C = d dx x 2 +1 ( ) 1/2 + C = 1 2 x 2...

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Unformatted text preview: d dx x 2 +1 + C = d dx x 2 +1 ( ) 1/2 + C = 1 2 x 2 +1 ( ) ¡ 1/2 ¢ 2 x = x x 2 +1 d dx x sin x +cos x + C = x cos x + sin x ( ) ¢ 1 ¡ sin x = x cos x d dx x a 2 a 2 ¡ x 2 + C = 1 a 2 a 2 ¡ x 2 ¡ x ( ¡ x ) a 2 ¡ x 2 / a 2 ¡ x 2 = 1 a 2 a 2 ¡ x 2 ( ) + x 2 a 2 ¡ x 2 ( ) 3/2 = 1 a 2 ¡ x 2 ( ) 3 d dx ¡ x 2 + a 2 a 2 x + C = ¡ 1 a 2 d dx x 2 + a 2 x = ¡ x x / x 2 + a 2 ( ) ¡ x 2 + a 2 ¢ 1 a 2 x 2 = ¡ x 2 ¡ x 2 + a 2 ( ) a 2 x 2 x 2 + a 2 = 1 x 2 x 2 + a 2 £ x ¡ 3/4 dx = x ¡ 3/4+1 ¡ 3/4+1 + C = x 1/4 1/4 + C =4 x 1/4 + C £ 3 x dx = £ x 1/3 dx = x 4/3 4/3 + C = 3 4 x 4/3 + C £ x 3 +6 x +1 ( ) dx = x 4 4 +6 x 2 2 + x + C = 1 4 x 4 +3 x 2 + x + C £ x 1+2 x 4 ( ) dx = £ x +2 x 5 ( ) dx = x 2 2 +2 x 6 6 + C = 1 2 x 2 + 1 3 x 6 + C £ (1 ¡ t ) 2+ t 2 ( ) dt = £ 2 ¡ 2 t + t 2 ¡ t 3 ( ) dt =2 t ¡ 2 t 2 2 + t 3 3 ¡ t 4 4 + C =2 t ¡ t 2 + 1 3 t 3 ¡ 1 4 t 4 + C £ x 2 +1+ 1 x 2 +1 dx = x 3 3 + x +tan ¡ 1 x + C 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 1 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.4 Indefinite Integrals and the Net Change Theorem ¡ 2 ¢ x ( ) 2 dx = ¡ 4 ¢ 4 x + x ( ) dx =4 x ¢ 4 x 3/2 3/2 + x 2 2 + C =4 x ¢ 8 3 x 3/2 + 1 2 x 2 + C ¡ 3 e u + sec 2 u ( ) du = 3 e u + tan u + C ¡ sin x 1 ¢ sin 2 x dx = ¡ sin x cos 2 x dx = ¡ 1 cos x £ sin x cos x dx = ¡ sec x tan xdx =sec x + C ¡ sin 2 x sin x dx = ¡ 2sin x cos x sin x dx = ¡ 2cos xdx =2sin x + C ¡ x x dx = ¡ x 3/2 dx = 2 5 x 5/2 + C C =5 3 ¢ 2 ¢ 4 ¡ (cos x ¢ 2sin x ) dx =sin x +2cos x + C C =5 3 ¢ 2 ¢ 4 ¡ 2 (6 x 2 ¢ 4 x +5) dx = 6 £ 1 3 x 3 ¢ 4 £ 1 2 x 2 +5 x 2 = 2 x 3 ¢ 2 x 2 +5 x 2 =(16 ¢ 8+10) ¢ 0=18 ¡ 3 1 (1+2 x ¢ 4 x 3 ) dx = x +2 £ 1 2 x 2 ¢ 4 £ 1 4 x 4 3 1 = x + x 2 ¢ x 4 3 1 =(3+9 ¢ 81) ¢ (1+1 ¢ 1)= ¢ 69 ¢ 1= ¢ 70 ¡ ¢ 1 2 x ¢ e x ( ) dx = x 2 ¢ e x ¢ 1 = 0 ¢ 1 ( ) ¢ 1 ¢ e ¢ 1 ( ) = ¢ 2+1/ e 11. 12. 13. 14. 15. . The members of the family in the figure correspond to , , , , and . 16. . The members of the family in the figure correspond to , , , , and . 17. 18. 19. 2 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.4 Indefinite Integrals and the Net Change Theorem ¡ ¢ 2 ( u 5 ¢ u 3 + u 2 ) du = 1 6 u 6 ¢ 1 4 u 4 + 1 3 u 3 ¢ 2 =0 ¢ 32 3 ¢ 4 ¢ 8 3 = ¢ 4 ¡ 2 ¢ 2 (3 u +1) 2 du = ¡ 2 ¢ 2 9 u 2 +6 u +1 ( ) du = 9 £ 1 3 u 3 +6 £ 1 2 u 2 + u 2 ¢ 2 = 3 u 3 +3 u 2 + u 2 ¢ 2 =(24+12+2) ¢ ( ¢ 24+12 ¢ 2)=38 ¢ ( ¢ 14)=52 ¡ 4 (2 v +5)(3 v ¢ 1) dv = ¡ 4 (6 v 2 +13 v ¢ 5) dv = 6 £ 1 3 v 3 +13 £ 1 2 v 2 ¢ 5 v 4 = 2 v 3 + 13 2 v 2 ¢ 5 v 4 =(128+104 ¢ 20) ¢ 0=212 ¡ 4 1 t (1+ t ) dt = ¡ 4 1 ( t 1/2 + t 3/2 ) dt = 2 3 t 3/2 + 2 5 t 5/2 4 1 = 16 3 + 64 5 ¢ 2 3 + 2 5 = 14 3 + 62 5 = 256 15 ¡ 9 2 t dt = ¡ 9 2 t 1/2 dt = 2 £ 2 3 t 3/2 9 = 2 £ 2 3 £ 27 ¢ 0=18 2 ¡ ¢ 1 ¢...
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5.4 - d dx x 2 +1 + C = d dx x 2 +1 ( ) 1/2 + C = 1 2 x 2...

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