Integration 227
=
1
2
∫
(u−1)u^1 /^2 du
=
1
2
∫
(u^3 /^2 −u^1 /^2 )du
=
1
2
(
u^5 /^2
5 / 2
−
u^3 /^2
3 / 2
)
+C
=
u^5 /^2
5
−
u^3 /^2
3
+C
=
(x^2 +1)^5 /^2
5
−
(x^2 +1)^3 /^2
3
+C.
- Letu=x−1;du=dxand (u+1)=x.
Rewrite:
∫
(u+1)^2 + 5
√
u
du
=
∫
u^2 + 2 u+ 6
u^1 /^2
du
=
∫ (
u^3 /^2 + 2 u^1 /^2 + 6 u−^1 /^2
)
du
=
u^5 /^2
5 / 2
+
2 u^3 /^2
3 / 2
+
6 u^1 /^2
1 / 2
+C
=
2(x−1)^5 /^2
5
+
4(x−1)^3 /^2
3
+12(x−1)^1 /^2 +C.
- Letu=
x
2
;du=
1
2
dxor 2du=dx.
Rewrite:
∫
tanu(2du)= 2
∫
tanudu
=−2ln|cosu|+C
=−2ln|cos
x
2
|+C.
- Letu=x^2 ;du= 2 xdxor
du
2
=xdx.
Rewrite:
∫
csc^2 u
du
2
=
1
2
∫
csc^2 udu
=−
1
2
cotu+C
=−
1
2
cot(x^2 )+C.
- Letu=cosx;du=−sinxdxor
−du=sinxdx.
Rewrite:
∫
−du
u^3
=−
∫
du
u^3
=−
u−^2
− 2
+C=
1
2 cos^2 x
+C.
9.Rewrite:
∫
1
(x^2 + 2 x+ 1 )+ 9
dx
=
∫
1
(x+1)^2 + 32
dx.
Letu=x+1;du=dx.
Rewrite:
∫
1
u^2 + 32
du
=
1
3
tan−^1
(
u
3
)
+C
=
1
3
tan−^1
(
x+ 1
3
)
+C.
- 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.
