4 Apr 2016
School
Department
Course
Professor

xndx
xn+1
n+1
(
1
x
)
dx
ln
|
x
|
+c
exdx
ex
!"
#
4x3dx=x4+c
x
(¿¿4+c)'=(x4)'+(c)'
¿
4x3
(x4)
4x3
¿4x3
!
$
%
3t4dt
3t4dt
"
t4+1
4+1+c
#
xn+1
n+1
find more resources at oneclass.com
find more resources at oneclass.com