dy
ln x 2
.
=
dx
x
ln x 2
Solution. y = ∫
dx . There are two cases to consider if the logarithm property ln ab = b ln a is
x
2 ln x
used. First, if x > 0 , then y = ∫
dx . To carry out the integration, let u = ln x ; then
x
1
du = dx . Thus,
x
22. Solve
y =
Second, if x < 0 , then y =
∫
2
2u du = u 2 + C = ( ln x ) + C .
∫
2
ln x 2
dx =
x
∫
ln ( −x )
x
du
1 d
1
dx
Then
and so du =
.
=
⋅ ( −x ) =
x
dx
−x dx
x
Therefore, in both cases, solutions are given by
dx =
∫
2 ln ( −x )
x
Thus, y =
Now let u = ln ( - x ) .
dx .
∫ 2u du = u 2 + C = ( ln ( −x ))
2
+C .
y = ( ln x ) 2 + C .
(
)
( )
Note. An alternative substitution is v = ln ( x 2 ) . Then, dv = 2x x 2 dx = 2 x dx and
⎡ ln x 2 ⎤ 2
⎡ 2 ln x ⎤ 2
1
u2
⎢⎣
⎥⎦
⎦ + C = ( ln x
y = ∫ u du =
+C =
+C = ⎣
2
4
4
4
( )
28. Solve
2
)
+C .
dy
x
.
= e2x sin
dx
3
( )
Solution. The integral y =
⎛x ⎞
∫ e 2x sin ⎜⎜⎜⎝ 3 ⎟⎟⎟⎟⎠ dx
is found by integrating by parts. From the table
d ⎡ ⎤
⋅
dx ⎣ ⎦
u = sin
∫ ⎡⎣ ⋅ ⎤⎦ dx
( x3 )
1
x
cos
3
3
( )
-
we have
⎛x ⎞
( )
⎛x ⎞
e 2x dx = dv
–1
1 2x
e
2
+1
1
x
sin
9
3
1
+1
1
⎛x ⎞
1 2x
e
4
⎛x ⎞
1
∫ e 2x sin ⎜⎜⎜⎝ 3 ⎟⎟⎟⎟⎠ dx = 2 e 2x sin ⎜⎜⎜⎝ 3 ⎟⎟⎟⎟⎠ − 12 e 2x cos ⎜⎜⎜⎝ 3 ⎟⎟⎟⎟⎠ − 36 ∫ e 2x sin ⎜⎜⎜⎝ 3 ⎟⎟⎟⎟⎠ dx .
Solving for the integral,
we obtain
y=
⎛x ⎞
18
⎛x ⎞
3
⎛x ⎞
∫ e 2x sin ⎜⎜⎜⎝ 3 ⎟⎟⎟⎟⎠ dx = 37 e 2x sin ⎜⎜⎜⎝ 3 ⎟⎟⎟⎟⎠ − 37 e 2x cos ⎜⎜⎜⎝ 3 ⎟⎟⎟⎟⎠ + C .
1
2
Error! Reference source not found.
30. Solve
dy
x
.
= 2
dx
x - 5x + 6
Solution.
y=
x
∫ x 2 − 5x + 6 dx .
Since the substitution method will not work here, we expand the
integrand into the sum of two partial fractions:
x
A
B
=
+
( x − 2 )( x − 3 ) x − 2 x − 3
⇒ A( x − 3 ) + B ( x − 2 ) = x .
Setting x = 2 , we obtain A = - 2 . And x = 3 gives B = 3 . Thus,
1
3
y = −2∫
dx + 3∫
dx .
x −2
x −3
From this, we have y = 3 ln x - 3 - 2 ln x - 2 + C .
32. Solve
dy
1
.
= 3
dx
x + 9x
Solution.
y=
1
∫ x 3 + 9x dx .
Integrating using the method of partial fractions, we have
1
A Bx + C
;
= + 2
x
x + 9x
x +9
A( x 2 + 9 ) + ( Bx + C )x = 1;
3
⎧ A+B = 0
⎪
⎪
⎪
Thus, ⎪
C =0
⎨
⎪
⎪
9A = 1
⎪
⎪
⎩
y=
⎫
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎭
1
⇒ A= ;
9
1
1
∫ x 3 + 9x dx = 9 ∫
1
B = −A = − ;
9
C = 0 . Therefore,
dx 1
x dx
1
1
− ∫
= ln x − ln x 2 + 9 + C , or
x
9 x2 + 9 9
18
(
y=
34. Solve
( A + B ) x 2 + Cx + 9A = 1
)
1 ⎛⎜ x 2 ⎞⎟⎟
ln ⎜
⎟ +C .
18 ⎜⎜⎝ x 2 + 9 ⎟⎠
dy
= sec ( 3x ) tan ( 3x ) .
dx
Solution. Integrating with respect to x , we have y = ∫ sec (3x ) tan (3x ) dx .
Let u = 3x .
du = 3 dx . Thus,
y=
36. Solve
1
1
secu tan u du = sec (3x ) +C .
∫
3
3
dy
= 5x 2 1 + 4x 3 .
dx
Solution. Integrating, we have y = ∫ 5x 2 1 + 4x 3 dx . Let u = 1 + 4x 3 . Then du = 12x 2dx .
y=
38. Solve y =
x2 − 5
∫ x 3 − 3x 2 + 4x − 12 dx .
32
5
(1 + 4x 3 ) + C
18
Then
3
Error! Reference source not found.
Solution. The partial fraction expansion of the integrand is
x2 + 5
A
Bx + C
.
=
+ 2
x- 3
( x - 3 )( x 2 + 4 )
x +4
Clearing the denominators and grouping like terms together gives
( A + B ) x 2 + (C - 3B ) x + 4A - 3C = x 2 - 5 .
So, we want to find values for A, B, and C satisfying the equations
⎪⎧⎪ A + B = 1
⎪⎪
⎨ C − 3B = 0
⎪⎪
⎪⎪⎩ 4A − 3C = −5.
Solving, we find A =
y=
=
41. To solve
4
9
27
. Thus,
, B=
, C =
13
13
13
x2 − 5
4
dx
9
x dx
27
dx
∫ x 3 − 3x 2 + 4x − 12 dx = 13 ∫ x − 3 + 13 ∫ x 2 + 4 + 13 ∫ x 2 + 4
⎛x ⎞
4
9
27
ln x − 3 + ln x 2 + 4 + tan−1 ⎜⎜ ⎟⎟⎟ + K.
⎜⎝ 2 ⎟⎠
13
26
26
∂z
x
, integrate with respect to y:
=5+
∂y
4 + y2
z=
⎛
x
⎞⎟
1
⎛y ⎞
∫ ⎜⎜⎜⎝ 5 + 4 + y 2 ⎟⎟⎟⎠ dy + f ( y ) = 5y + 2 tan−1 ⎜⎜⎜⎝ 2 ⎟⎟⎟⎟⎠ + f ( x ) .
where f denotes a differentiable function of x.
The solution with Maple 11 is
> PDE1 := diff(z(x, y), y) = 5 + x/(4 + y^2);
> pdsolve(PDE1);
42. To solve zx = 2y - 5y 3 csc2 x , integrate with respect to x:
z=
∫ ( 2y − 5y 3 csc2 x )dx = 2xy + 5y 3 cotx + g ( y ) ,
where g denotes a differentiable function of y.
The solution with Maple 11 is
> PDE2 := diff(z(x, y), x) = 2*y - 5*y^3*csc(x)^2;
> pdsolve(PDE2);