MA 3972-MA-Book May 8, 2018 13:52
Integration 245
- Letu=
1
x
;du=
− 1
x^2
dxor−du
=
1
x^2
dx.
Rewrite:
∫
sec^2 u(−du)=−
∫
sec^2 udu
=−tanu+C
=−tan
(
1
x
)
+C.
- Rewrite:
∫
e(2x+^4 x)dx=
∫
e^6 xdx.
Letu= 6 x;du= 6 dxor
du
6
=dx.
Rewrite:
∫
eu
du
6
=
1
6
∫
eudu
=
1
6
eu+C=
1
6
e^6 x+C.
- Letu=lnx;du=
1
x
dx.
Rewrite:
∫
1
u
du=ln|u|+C
=ln|lnx|+C.
- Sinceexand lnxare inverse functions:
∫
ln
(
e^5 x+^1
)
dx=
∫
(5x+1)dx
=
5 x^2
2
+x+C.
- Rewrite:
∫ (
e^4 x
ex
−
1
ex
)
dx
=
∫ (
e^3 x−e−x
)
dx
=
∫
e^3 xdx−
∫
e−xdx.
Letu= 3 x;du= 3 dx;
∫
e^3 xdx=
∫
eu
(
du
3
)
=
1
3
eu+C 1
=
1
3
e^3 x+C.
Letv=−x;dv=−dx;
∫
e−xdx=
∫
ev(−dv)=ev+C 2
=−e−x+C 2
Thus,
∫
e^3 xdx−
∫
e−xdx
=
1
3
e^3 x+e−x+C.
Note thatC 1 and C 2 are arbitrary
constants, and thus,C 1 +C 2 =C.
- Rewrite:∫
(9−x^2 )x^1 /^2 dx=
∫ (
9 x^1 /^2 −x^5 /^2
)
dx
=
9 x^3 /^2
3 / 2
−
x^7 /^2
7 / 2
+C
= 6 x^3 /^2 −
2 x^7 /^2
7
+C.
- Letu= 1 +x^3 /^2 ;du=
3
2
x^1 /^2 dxor
2
3
du=x^1 /^2 dx=
√
xdx.
Rewrite:
∫
u^4
(
2
3
du
)
=
2
3
∫
u^4 du
=
2
3
(
u^5
5
)
+C=
2
(
1 +x^3 /^2
)
15
5
+C.
- Since
dy
dx
=ex+2, theny=
∫
(ex+ 2 )dx=ex+ 2 x+C.
The point (0, 6) is on the graph ofy.
Thus, 6=e^0 +2(0)+C⇒ 6 = 1 +Cor
C=5. Therefore,y=ex+ 2 x+5.
- Letu=ex;du=exdx.
Rewrite:− 3
∫
sin(u)du=−3(−cosu)+C
=3 cos(ex)+C.
- Letu=ex+e−x;du=(ex−e−x)dx.
Rewrite:
∫
1
u
du=ln|u|+C
=ln|ex+e−x|+C
or=ln
∣∣
∣∣ex+^1
ex
∣∣
∣∣+C
=ln
∣∣
∣∣e
2 x+ 1
ex
∣∣
∣∣+C
=ln|e^2 x+ 1 |−ln|ex|+C
=ln|e^2 x+ 1 |−x+C.
