Ad. 1
I =
Z
x ln |x| dx
(x
2
+ 1)
2
=
u = ln |x| v
0
=
x
(x
2
+ 1)
2
u
0
=
1
x
v = −
1
2
1
x
2
+ 1
= −
ln |x|
2(x
2
+ 1)
+
Z
1
2x(x
2
+ 1)
dx
1
2x(x
2
+ 1)
=
(1 + x
2
) − x
2
2x(x
2
+ 1)
=
1
2x
−
x
2(x
2
+ 1)
I = −
ln |x|
2(x
2
+ 1)
+
1
2
ln |x| −
1
4
ln(x
2
+ 1) + C =
x
2
ln |x|
2(x
2
+ 1)
−
1
4
ln(x
2
+ 1) + C
I =
Z
ln |x| dx
(x + 2)
2
=
u = ln |x| v
0
=
1
(x + 2)
2
u
0
=
1
x
v = −
1
x + 2
= −
ln |x|
x + 2
+
Z
1
x(x + 2)
dx =
= −
ln |x|
x + 2
+
Z
1
2
1
x
−
1
x + 2
dx = −
ln |x|
x + 2
+
1
2
ln |x| −
1
2
ln |x + 2| + C = −
ln |x|
x + 2
+
1
2
ln
x
x + 2
+ C
lub: =
−
1
x + 2
+
1
2
ln |x| −
1
2
ln |x + 2| + C =
x ln |x|
2(x + 2)
−
1
2
ln |x + 2| + C
I =
Z
arc tg
√
x dx ==
u = arc tg
√
x
v
0
= 1
u
0
=
1
1 + x
·
1
2
√
x
v = x
= x arc tg
√
x −
Z
√
x
2(1 + x)
dx =
=
√x = t, x = t
2
, dx = 2t dt
= x arc tg
√
x −
Z
t · 2t dt
2(1 + t
2
)
= x arc tg
√
x −
Z
t
2
dt
(1 + t
2
)
Ponieważ
t
2
(1 + t
2
)
=
t
2
+ 1 − 1
t
2
+ 1
= 1 −
1
t
2
+ 1
więc I = x arc tg
√
x − t + arc tg t + C = arc tg
√
x −
√
x + arc tg
√
x = (x + 1) arc tg
√
x −
√
x + C
I =
Z
x
2
ln(x
2
+x+2) dx =
(u = ln(x
2
+ x + 2) v
0
= x
2
u
0
=
2x + 1
x
2
+ x + 2
v =
1
3
x
3
)
=
1
3
x
3
ln(x
2
+x+2)−
Z
1
3
2x
4
+ x
3
x
2
+ x + 2
dx
Dzielenie daje:
2x
2
− x − 3
x
2
+ x + 2
2x
4
+ x
3
− 2x
4
− 2x
3
− 4x
2
− x
3
− 4x
2
x
3
+ x
2
+ 2x
− 3x
2
+ 2x
3x
2
+ 3x + 6
5x + 6
Ad. 2
Z
1
x
2
r
x + 1
x
dx =
x + 1
x
= t
2
x =
1
t
2
− 1
x + 1 = xt
2
dx =
−2t
(t
2
− 1)
2
dt
x(1 − t
2
) = −1
= −
Z
(t
2
−1)
2
t
2t
(t
2
− 1)
2
dt = −
Z
2t
2
dt =
= −
2
3
t
3
+ C = −
2
3
x + 1
x
3/2
+ C
Ewentualnie, podstawienie u =
x + 1
x
= 1 +
1
x
, du = −
1
x
2
dx.
Z
1
x
r
x − 2
x
dx =
x − 2
x
= t
2
x = −
2
t
2
− 1
x − 2 = xt
2
dx =
4t
(t
2
− 1)
2
dt
x(1 − t
2
) = 2
=
Z
−
t
2
− 1
2
t
4t
(t
2
− 1)
2
dt = −
Z
2t
2
t
2
− 1
dt =
= −
Z
(2t
2
− 2) + 2
t
2
− 1
dt = −
Z
2 +
1
t − 1
−
1
t + 1
dt = −
2t + ln
t − 1
t + 1
+ C =
= −2
r
x − 2
x
− ln
q
x−2
x
− 1
q
x−2
x
+ 1
+ C = −2
r
x − 2
x
− ln
q
x−2
x
− 1
2
x−2
x
− 1
=
= −2
r
x − 2
x
− 2 ln
r
x − 2
x
− 1
+ ln 2 − ln |x| + C = −2
r
x − 2
x
− 2 ln
r
x − 2
x
− 1
− ln |x| + e
C
Z
dx
x − 1 −
√
x − 1
= (
√
x − 1 = t, x − 1 = t
2
, x = t
2
+ 1, dx = 2t dt) =
Z
2t dt
t
2
− t
=
Z
2
t − 1
dt =
= 2 ln |t − 1| + C = 2 ln
√
x − 1 − 1
+ C
I =
Z
arc tg 1 +
√
x
dx = (1 +
√
x = t,
√
x = t − 1, x = (t − 1)
2
, dx = 2(t − 1) dt)
=
Z
arc tg t · 2(t − 1) dt =
(u = arc tg t v
0
= 2t − 2
u
0
=
1
1 + t
2
v = t
2
− 2t
)
= (t
2
− 2t) arc tg t −
Z
t
2
− 2t
t
2
+ 1
dt
t
2
− 2t
t
2
+ 1
=
t
2
+ 1 − 2t − 1
t
2
+ 1
= 1 −
2t + 1
t
2
+ 1
I = (t
2
− 2t) arc tg t − t + ln(t
2
+ 1) + arc tg t + C
a ponieważ t
2
− 2t = 1 + 2
√
x + x − 2 − 2
√
x = x − 1 i t
2
+ 1 = x + 2
√
x + 1 + 1 = t
2
+ 2
√
x + 2
więc I = (x − 1) arc tg 1 +
√
x
− 1 −
√
x + ln x + 2
√
x + 2
+ arc tg 1 +
√
x
=
= x arc tg 1 +
√
x
−
√
x + ln x + 2
√
x + 2
+
e
C
Ad. 4
I =
Z
dx
sin
4
x + cos
4
x
sin
4
x + cos
4
x = (sin
2
x + cos
2
x)
2
− 2 sin
2
x cos
2
x = 1 −
1
2
sin
2
x cos
2
x;
I =
Z
dx
1 −
1
2
sin
2
2x
=
Z
2 dx
2 − sin
2
2x
=
=
tg x = t; sin 2x =
2t
1 + t
2
; x = arc tg t + kπ; dx =
1
1 + t
2
dt
=
=
Z
1
2 −
4t
2
1+t
2
·
1
1 + t
2
dt =
Z
1 + t
2
2(1 + t
2
)
2
− 4t
2
dt =
Z
t
2
+ 1
2(t
4
+ 1)
dt
t
2
+ 1
2(t
4
+ 1)
=
At + B
t
2
−
√
2t + 1
+
Ct + D
t
2
+
√
2t + 1
t
2
+ 1 = 2(At + B)(t
2
+
√
2t + 1) + 2(Ct + D)(t
2
−
√
2t + 1)
t
2
+ 1 =
2At
3
+2Bt
2
+2A
√
2t
2
+2B
√
2t+2At+2B+
+ 2Ct
3
+2Dt
2
−2C
√
2t
2
−2D
√
2t+2Ct+2D
t
3
:
2A
+ 2C
= 0
t
2
:
2A
√
2+2B
− 2C
√
2 + 2D
= 1
t :
2A
+2B
√
2 + 2C
− 2D
√
2 = 0
1 :
2B
+ 2D
= 1
Z pierwszego i trzeciego równania B = D, więc na mocy czwartego B = D =
1
4
.
Teraz z drugiego A − C = 0, co w połączeniu z pierwszym daje A = C = 0. Zatem
I =
Z
1
4(t
2
−
√
2t + 1)
+
1
4(t
2
+
√
2t + 1)
dt =
1
4
2
√
2
arc tg
2t +
√
2
√
2
+
1
4
2
√
2
arc tg
2t −
√
2
√
2
+ C =
=
1
2
√
2
arc tg
√
2t + 1
+
1
2
√
2
arc tg
√
2t − 1
+ C =
=
1
2
√
2
arc tg
√
2 tg x + 1
+
1
2
√
2
arc tg
√
2 tg x − 1
+ C
Ad. 7
Z
dx
1 + sin x + cos x
=
t = tg
x
2
; x = 2 arc tg t + 2kπ; dx =
2
1 + t
2
dt; sin x =
2t
1 + t
2
; cos x =
1 − t
2
1 + t
2
;
tg
x
2
=
1 − cos x
sin x
=
sin x
1 + cos x
=
Z
2
1+t
2
dt
1 +
2t
1+t
2
+
1−t
2
1+t
2
=
Z
2
1 + t
2
+ 2t + 1 − t
2
dt =
Z
2
2t + 2
dt =
=
Z
1
t + 1
dt = ln |t + 1| + C = ln
tg
x
2
+ 1
+ C = ln
1 − cos x
sin x
+ 1
+ C = ln
sin x − cos x + 1
sin x
+ C
lub: = ln
sin x
1 + cos x
+ 1
+ C = ln
sin x + cos x + 1
1 + cos x
+ C
