MA 3972-MA-Book May 8, 2018 13:52244 STEP 4. Review the Knowledge You Need to Score High
11.6 Solutions to Practice Problems
1.
x^6
6
+x^3 −
x^2
2
+x+C- Rewrite:
∫
(x^1 /^2 −x−^2 )dx=
x^3 /^2
3 / 2−
x−^1
− 1+C
=
2 x^3 /^2
3+
1
x+C.
- Letu=x^4 −10;du= 4 x^3 dxor
du
4
=x^3 dx.
Rewrite:∫
u^5
du
4=
1
4
∫
u^5 du=(
1
4)
u^6
6+C
=
(x^4 −10)^6
24+C.
- Letu=x^2 + 1 ⇒(u−1)=x^2 and
du= 2 xdxor
du
2
=xdx.
Rewrite:∫
x^2√
x^2 +1(xdx)=∫
(u−1)√
u
du
2=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.
- Rewrite:
∫
1
(x^2 + 2 x+ 1 )+ 9
dx=
∫
1
(x+1)^2 + 32dx.Letu=x+1;du=dx.
Rewrite:∫
1
u^2 + 32du=
1
3
tan−^1(
u
3)
+C=
1
3
tan−^1(
x+ 1
3)
+C.