Uttryck 12: left bracket, StartFraction, 1 Over "x" , EndFraction "u" squared times "x" "d" "u" , right bracket equals left bracket, "u" squared "d" "u" , right bracket1xu2·xdu=u2du
12
Nu kan integralet bestemmes:
13
Uttryck 14: left bracket, "u" squared "d" "u" , right bracket equals StartFraction, "u" cubed Over 3 , EndFraction plus "c"u2du=u33+c
14
Dermed er stamfunktionen
15
Uttryck 16: "F" left parenthesis, "x" , right parenthesis equals StartFraction, left parenthesis, ln "x" plus 5 , right parenthesis cubed Over 3 , EndFraction plus "c"Fx=lnx+533+c
16
Uttryck 17: Start integral from "i" to "x" , end integral, StartFraction, left parenthesis, ln "t" plus 5 , right parenthesis squared Over "t" , EndFraction "d" "t" plus "c"∫xilnt+52tdt+c
