B. Jurˇciˇc Zlobec1
1Univerza v Ljubljani, Fakulteta za Elektrotehniko 1000 Ljubljana, Trˇzaˇska 25, Slovenija
Matematika FE, Ljubljana, Slovenija 4. januar 2013
I Linearnost:
Z
(a f(x) +b g(x))dx =a Z
f(x)dx +b Z
g(x)dx.
II Per partes:
Z
f0(x)g(x)dx =f(x)g(x)− Z
f(x)g0(x)dx.
III Substitucija:
Z
f(g(x))g0(x)dx =F(g(x)), kjer je Rf(x)dx =F(x).
u=g(x)→,du=g0(x)dx →,R
f(u)du =F(u) =F(g(x)).
IV Linearna substitucija:R
f(ax +b)dx = 1aF(ax+b).
1. R
xndx = xn+1n+1, n∈Z\ {−1}, x 6=0.
2. R
xsdx = xs+1s+1, s∈R\ {−1}, x >0.
3. R
exdx =ex. 4. R dx
x =ln|x|.
5. R
sinx dx=−cosx. 6. R
cosx dx =sinx. 7. R dx
a2+x2 = 1aarctanxa, a>0 . 8. R √dx
a2−x2 =arcsinxa, a>0.
9. R √dx
x2±a2 =ln(x +√
x2±a2), a>0.
A Integral oblike Z
f(x)dx =
Z P(x)dx
(x−a)n(x2+px +q)m, kjer jem,n∈N,p2−4q <0 inP(x)polinom stopnje manj ali enako 2m+n−1, reˇsimo s pomoˇcjo nastavka:
Z
f(x)dx = Q2m−2+n−2(x)
(x −a)n−1(x2+px +q)m−1+ +Aln|x−a|+Bln(x2+px+q) +Carctg√2x+p
4q−p2. B Integral oblike
Z
f(x)dx =
Z Pn(x)dx px2+px+q reˇsimo s pomoˇcjo nastavka:
Z
f(x)dx =Qn−1(x)p
x2+px+q+λ
Z dx
px2+px +q.
f(x) =3+ 1 x + 1
x2.
I Uporabimo (I)→(1)→(4).
I R
3+x1+x12
dx =→
I R
3dx+R dx
x +R dx
x2 dx =→
I
Z 3+ 1
x + 1 x2
dx =3x+ln|x| − 1 x +C.
f(x) = 3+2x2 1+x2 .
I Uporabimo (I)→(1)→(7).
I R 3+2x2
1+x2 dx =R 1+2(1+x2)
1+x2 dx →
I R
2dx+R dx
1+x2 =→
I
Z 3+2x2
1+x2 dx =2x+arctgx+C.
f(x) = x3 1+x2.
I Preverimo stopnji ˇstevca in imenovalca.
I Delimo: 1+xx32 =x− 1+xx 2
I (I)→(1)→(III)→(4).
I u =1+x2,du =2x dx →
I 1
2x2−12R du
u = 12x2− 12ln|u| →
I 1
2x2−12ln(1+x2)→
I
Z x3 1+x2 = 1
2x2−lnp
1+x2+C
f(x) = 3
2+2x+x2 + x 1+x2.
I Uporabimo (I)→(IV)→(III)→(7)→(4).
I R
3dx
2+2x+x2 +1+xx 2
dx =3R dx
(x+1)2+1+R x
1+x2dx →
I u =x+1,du=dx, v =1+x2,dv =2x dx →
I 3R du
1+u2 + 12R dv
v =→
I
Z
f(x)dx =3 arctg(x+1) +1
2ln(1+x2) +C.
f(x) = x−1 x(x +1).
I Razcepimo na parcialne ulomke.
x−1
x(x+1) = Ax + x+1B =−x1+x+12 .
I R x−1
x(x+1)dx =−R dx
x +R 2dx
x+1.
I R x−1
x(x+1)dx =−ln|x|+2 ln|x+1| →
I
Z x −1
x(x +1)dx =ln(x +1)2
|x| +C.
f(x) = x2+1 x(1−x2).
I Razcepimo na parcialne ulomke.
x2+1
x(1−x2) = Ax +1−xB + 1+xC →
I x2+1=A(1−x)(1+x) +Bx(1+x) +Cx(1−x)→
I −A+B−C=1,B+C=0 inA=1.
I
Z x2+1
x(1−x2)dx =ln|x| −ln|1−x2|.
f(x) = x −1 x(x2+1).
I Razcepimo na parcialne ulomke.
x−1
x(x2+1) = Ax +Bx+Cx2+1 =−1x +x1+x2+1.
I R x−1
(x2+1)dx =−R dx
x +R dx
1+x2 +R x dx
1+x2.
I R x−1
x(x+1)dx =−ln|x|+arctgx +12ln(x2+1)→
I
Z x −1
x(x +1)dx =−ln|x|+arctgx +lnp
1+x2+C.
f(x) = x −1 x2(x+1).
I Razcepimo na parcialne ulomke.
x−1
x2(x+1) = xA2 +Bx +x+1C =−x12 +x2−x+12 .
I R x−1
x(x+1)dx =−R dx
x2 +R 2dx
x −R 2dx
1+x.
I R x−1
x(x+1)dx = 1x +2 ln|x| −2 ln|x+1| →
I
Z x −1
x(x +1)dx = 1
x +ln x2
(1+x)2+C.
f(x) = x2+1 x2(1+x).
I Zapiˇsemo nastavek (A).
I R x2+1
x2(1+x)dx = Ax +Bln|x|+Cln|1+x| →
I x2+1=−A(1+x) +Bx(1+x) +Cx2→
I B+C=1,A+B=0 inA=−1.
I
Z x2+1
x2(1+x)dx =−1
x −ln|x|+2 ln|1+x|.
f(x) = x3−1 x2(1+x).
I Stopnji ˇstevca in imenovalca sta enaki, delimo.
x3−1
x2(1+x) =1− x2+1
x2(x+1).
I Zapiˇsemo nastavek (A).
R x2+1
x2(1+x)dx = Ax +Bln|x|+Cln|1+x| →
I x2+1=−A(1+x) +Bx(1+x) +Cx2→
I B+C=1,A+B=0 inA=−1.
I
Z
1− x2+1 x2(1+x)
dx =x+ 1
x +ln|x| −ln(1+x)2+C.
f(x) = x2−1 x(2+2x+x2).
I Preverimo stopnji ˇstevca in imenovalca.
I Zapiˇsemo nastavek (A).
R x2−1
x(2+2x+x2)dx =Aln|x|+Bln|x2+x+1|+Carctg(1+x)→
I x2−1= Ax + 2Bx+B+Cx2+x+1 →
I B+C=1,A+B=0 inA=−1.
I
Z x2−1
x(x2+2x +2)dx =
−1
2ln|x| −1
2arctg(x+1) +3
4ln(2+2x+x2) +C.
f(x) = x
√x+1.
I (III)→(1).
I
Z x
√x+1dx →u =x+1,du =dx →
I R u−1
√u du=R
u12 −u−12 du.
I R x
√x+1dx = 23p
(x+1)3−2√
x +1→
I
Z x
√x+1dx = 2
3(x−2)√
1+x+C
f(x) =xp
1+x2.
I (III)→(2).
I
Z xp
1+x2→u=1+x2,du =2x dx →
I 1 2
R u12 du= 13u32.
I
Z xp
1+x2= 1 3
q
(1+x2)3+C.
f(x) =p
1−x2.
I (III)→(2).
I
Z p
1−x2dx →x =sinu,dx =cosu du→
I R p
1−sin2ucosu du=R
cos2u du=→
I u+12sin(2u) =u+sinucosu=u+sinup
1−sin2u.
I
Z p
1−x2dx = 1 2
arcsinx+xp 1−x2
+C.
f(x) =p
1−x2.
I Reˇsujemo s pomoˇcjo nastavka (B).
I
Z p
1−x2dx =→
I
Z 1−x2
√
1−x2dx = (Ax+B)p
1−x2+C
Z 1
√
1−x2dx →
I √1−x2
1−x2 =A√
1−x2−(Ax+B)x√
1−x2 +√C
1−x2 →
I 1−x2=−2Ax2−Bx+C+A→,A=C = 12,B=0.
I
Z p
1−x2dx = 1 2
arcsinx+xp
1−x2 +C.
f(x) =p
1+x2.
I Reˇsujemo s pomoˇcjo nastavka (B).
I
Z 1+x2
√
1+x2dx = (Ax+B)p
1+x2+C
Z 1
√
1+x2dx →
I √1+x2
1+x2 =A√
1+x2+(Ax+B)x√
1+x2 +√C
1+x2 →
I 1+x2=2Ax2+Bx +C+A→,A=C = 12,B =0.
I
Z p
1+x2dx = 1 2
ln(x+p
1+x2) +xp
1+x2 +C.
f(x) = x +2
√
2+2x +x2.
I Reˇsujemo s pomoˇcjo nastavka (B).
I
Z x+2
√2+2x+x2dx =
Ap
2+2x+x2+B
Z dx
√
2+2x+x2 →
I √x+2
1+x2 = +√A(x+1)
2+2x+x2 +√ B
2+2x+x2 →
I x +2=A(x +1) +B →,A=1,B=1.
I
Z √
x+2dx =
p2+2x+x2+ln(x+1+p
x2+2x+2) +C.
f(x) = x +2
√
1+2x −x2.
I Reˇsujemo s pomoˇcjo nastavka (B).
I
Z x+2
√
1+2x−x2dx = Ap
1+2x−x2+B
Z dx
√
1+2x−x2 →
I √ x+2
1+2x−x2 = A(2−2x)
2
√
1+2x−x2 +√ B
1+2x−x2 →
I x +2=A−Ax+B,A=−1 inB=3.
I R √ 1
1+2x−x2dx =R √ dx
2−(x−1)2 =→(IV)→(8)=arcsin(x−1).
I
Z x+2
√2x−x2dx =−p
1+2x −x2−3 arcsinx −1
√2 +C.
f(x) = 1 xlnx.
I (III)→(4).
I
Z dx
xlnx →u=lnx,du = dx x →
I R du
u =lnu→
I
Z dx
xlnx =ln|lnx|+C.
f(x) = ex 1+e2x.
I (III)→(7).
I
Z ex
1+e2x dx →
I u =ex,du =exdx →
I R du
1+u2 =arctgu→
I
Z ex
1+e2x dx =arctgex+C.
f(x) =tgx.
I Upoˇstevamo, da je tgx = sinx cosx →
I (III)→(4).
I u =cosx,dx =−sinx dx →
I
Z sinx cosx =
Z du
u =−ln|u| →
I
Z
tgx dx =−ln|cosx|+C
f(x) =cos2x.
I Upoˇstevamo, da je cos2x = 12(1+cos(2x))
I (I)→(1)→(IV)→(6).
I R
cos2x dx =R
dx +R
cos(2x)dx =→
I
Z
cos2x dx = x 2 +1
4sin(2x) +C.
f(x) =xe−x
I (II)→(1)→(IV)→(3).
I
Z
xe−xdx →u =x,dv =e−xdx, du=dx,v =−e−x →
I −xe−x+R
e−xdx =−xe−x−e−x
I
Z
xe−xdx =−e−x(x+1) +C
f(x) = sinx 1+cosx.
I
Z sinx
1+cosx dx →
I u =cosx,du =−sinx dx →
I −R du
1+u =−ln|1+u|=−ln|1+cosx| →
I
Z sinx
1+cosx dx =−ln|1+cosx|+C
f(x) =e−xcos(2x).
I Dvakrat uporabimo (II).
I R
e−xcos(2x)dx →
I u =e−x,dv =cos(2x)dx, du =−e−xdx,v = 12sin(2x)
I R
e−xcos(2x)dx = 12e−xsin(2x) + 12R
e−xsin(2x)dx
I 1 2
R e−xsin(2x)dx →
I u =e−x,dv =sin(2x)dx, du=−e−x,v =−12cos(2x)
I 1 2
R e−xsin(2x)dx =−14 e−xcos(2x) +R
e−xcos(2x)
I
Z
e−xcos(2x)dx = 1
5e−x(2 sin(2x)−cos(2x)) +C
f(x) =x2lnx.
I Uporabimo (II).
I
Z
x2lnx dx→
I u =lnx,dv =x2dx, du = dxx ,v = x33 →
I R
x2lnx dx = 13 x3lnx−R
x2dx .
I
Z
x2lnx dx= x3
3 lnx−x3 9 +C
S= Z 4
0
e
√xdx.
I Uvedemot =√
x →t2=x →2t dt =dx.
I Ker jex =0→t =0 inx =4→t=2,S=2R2 0 tetdt.
I Integriramoper partes:
u =t, dv =etdt →du=dt, v =et.
I S= 2tet
2 0−2R2
0 etdt =4e2−2e2−2=2(e2−1).
I
Z 4 0
e
√xdx =2(e2−1).
S= Z 2
0
xex2dx.
I Uvedemot =x2→dt=2xdx →dx = dt2.
I Ker jex =0→t =0 inx =2→t=4,S= 12R4 0 etdt.
I S= 1 2
Z 4 0
etdt = 1
2(e4−1).
S= Z π2
0
sin√ x dx.
I Uvedemot =√
x,→t2=x,→2tdt =dx.
I Ker jex =0→t =0 inx =π2→t =π, velja S=2Rπ
0 tsint dt.
I Integriramoper partes:
u =t, dv =sint dt,→du=dt, v =−cost.
I S= −2tcost|π0+2Rπ
0 cost dt =π S=2
Z π
0
tsint dt =2π.
S= Z π
4
0
dx cos2x.
I Uvedemot =tgx.
I dt = cosdx2x. Nove mejex =0→t=0,x = π4 →t =1.
I
Z 1 0
dt= t|10=1.
Z 9 4
√dx x −1.
I Uvedemo
x =t2,→dx =2t dt, x =4→t=2, x =9→t =3.
I 2R3 2
t dt
t−1 = t+ln|t−1||32=2+ln 4.
I
Z 9 4
√dx
x −1 =2+ln 4.
Z 2
−1
dx 9−x2.
I Razcepimo na parcialne ulomke: 9−x12 = 16 1
x−3+x+31
I R2
−1 dx 9−x2 =R2
−11 6
1
x−3+x+31
dx = 16ln|x2−3|
2
−1.
I
Z 2
−1
dx
9−x2 = ln 10 6 .
I Uporabimo (III)
→u =xn,dv =e−x, du=nxn−1,v =−e−x
I PiˇsimoΠ(n) =R∞
0 xne−xdx.
I R∞
0 xne−xdx =−xne−x|∞0 +nR∞
0 xn−1e−xdx
I Uporabimo l’H ˆospitalovo pravilo pri raˇcunanju limite limx→∞xne−x =limx→∞ xenx =0,
I in dobimo zvezoΠ(n) =nΠ(n−1).
I Ker jeΠ(0) =R∞
0 e−x = −e−x|∞0 =1,
I lahko z matematiˇcno indukcijo pokaˇzemo, da jeΠ(n) =n!
Z ∞
1
dx x2.
I
Z ∞
1
dx
x = −1 x
∞ 1
=1.
I
Z ∞
0
dx x2 =1.
Z 4 0
√dx 4−x.
I Uvedemot2=4−x →2t dt =−dx.
I Mejex =0→t =2, x =4→t =0.
I −2 Z 0
2
t dt
t = −2t|02=4
I
Z 4 0
√dx
4−x =4
Z 1
−2
|x| x dx.
I
Z 1
−2
|x|
x dx = Z 0
−2
(−1)dx+ Z 1
0
dx = −x|0−2+x|10=−1
I
Z 1
−2
|x|
x dx =−1
Z ∞
0
e−
√xdx.
I Uvedemot2=x →2t dt =dx.
I Mejex =0→t =0, x → ∞ →t→ ∞.
I R∞ 0 e−
√xdx =2R∞ 0 te−tdt
I Integriramoper partes:
du =e−tdx,v =t →u=−e−t,dv =dt.
I 2R∞
0 te−tdt = −2te−t
∞
0 +2R∞ 0 e−tdt
I Ker je limt→∞te−t =0, je 2R∞
0 te−tdt =2.
I
Z ∞
0
e−
√xdx =2.
Z ∞
0
e−
√xdx
√x .
I Uvedemot2=x →2t dt =dx.
I Mejex =0→t =0, x → ∞ →t→ ∞.
I
Z ∞
0
e−
√xdx
√x =2 Z ∞
0
e−tdt = −2e−t
∞ 0 =2.
I
Z ∞
0
e−
√xdx
√x =2.
Z 1 0
√xln1 x dx.
I R1 0
√xln1xdx =−R1 0
√xlnx dx
I Integriramoper partes.
u =lnx,dv =√
x dx →du = dxx ,v = 23
√ x3
I
Z 1 0
√xlnx dx = 2 3
√ x3lnx
1 0
−2 3
Z 1 0
√x dx
I
Z 1 0
√xln1
x dx = 4 9