52 1 Mathematical Physics
=−
1
5
ln(2x+1)+
1
5
ln(x−2)+C
=
1
5
ln
(
x− 2
2 x+ 1
)
+C
1.42r^2 =a^2 sin 2θ
Elementary area
dA=
1
2
r^2 dθ
A=
1
2
∫π/ 2
0
r^2 dθ=
a^2
2
∫π/ 2
0
sin 2θdθ=
a^2
∫π/ 2
0
sinθd(sinθ)=
a^2
2
sin^2 θ|^10 =
a^2
2
Fig. 1.11Polar diagram of
the curver^2 =a^2 sin 2θ
1.43 Sincex^2 +2 occurs twice as a factor, assume
x^3 +x^2 + 2
(x^2 +2)^2
=
Ax+B
(x^2 +2)^2
+
Cx+D
x^2 + 2
On clearing off the fractions, we get
x^3 +x^2 + 2 =Ax+B+(Cx+D)(x^2 +2)
orx^3 +x^2 + 2 =Cx^3 +Dx^2 +(A+ 2 C)x+B+ 2 D
Equating the coefficients of like powers ofx
C= 1 ,D= 1 ,A+ 2 C= 0 ,B+ 2 D= 2
This givesA=− 2 ,B= 0 ,C= 1 ,D= 1
Hence,
x^3 +x^2 + 2
(x^2 +2)^2
=−
2 x
(x^2 +2)^2
+
x
x^2 + 2
+
1
x^2 + 2
∫
(x^3 +x^2 +2)dx
(x^2 +2)^2
=−
∫
2 xdx
(x^2 +2)^2
+
∫
xdx
x^2 + 2
+
∫
dx
x^2 + 2
=
1
x^2 + 2
+
1
2
ln(x^2 +2)+
1
√
2
tan−^1
(
x
√
2
)
+C
1.44
∫∞
0
4 a^3 dx
x^2 + 4 a^2
=lim
b=∞
∫b
0
4 a^3 dx
x^2 + 4 a^2
=lim
b=∞
[
2 a^2 tan−^1
(x
2 a
)]b
0
=limb=∞
[
2 a^2 tan−^1
(
b
2 a
)]
= 2 a^2.
π
2
=πa^2