Integral Table Pdf : Ejercicios propuestos de Integrales, Método de Sustitución / ©2005 be shapiro page 3 this document may not be reproduced, posted or published without permission.
If a term in your choice for yp happens to be a solution of the homogeneous ode corresponding to (4), multiply this term by x (or by x 2 if this solution corresponds to a double root of the Bei integralen über echt gebrochenrationale funktionen wird auf die methode der partialbruchzerlegung verwiesen. Sn+1 (11) tx (x 1 2r) ( x+ 1) sx+1 (12) sinkt k s2 + k2 (13) coskt s s2 + k2 (14) eat 1 s a (15) sinhkt k. E−ax2dx= 1 2 π a # $% & '(1 2 0 ∞ ∫ ax xe−2dx= 1 2a 0 ∞ ∫ x2e−ax2dx= 1 4a π a # $% & '(1 2 0 ∞ ∫ x3e−ax2dx= 1 2a2 0 ∞ ∫ x2ne−ax2dx= 1⋅3⋅5⋅⋅⋅(2n−1) 2n+1an π a $ %& ' 1 2 0 ∞ ∫ x2n+1e−ax2dx= n! Udv a b ∫=#uv$% a b −vdu a b ∫ u and v are.
Z xn dx= xn+1 n+1 +c (n6= 1) 2.
If a term in your choice for yp happens to be a solution of the homogeneous ode corresponding to (4), multiply this term by x (or by x 2 if this solution corresponds to a double root of the Table of useful integrals, etc. N6= 1 (2) z 1 x dx= lnjxj (3) z udv= uv z vdu (4) z 1 ax+ b dx= 1 a lnjax+ bj integrals of rational functions (5) z 1 (x+ a)2 dx= 1 x+ a (6) z (x+ a)ndx= (x+ a)n+1 n+ 1;n6= 1 (7) z x(x+ a)ndx= (x+ a)n+1((n+ 1)x a) (n+ 1)(n+ 2) (8) z 1 1 + x2 dx= tan 1 x (9) z 1 a2 + x2 dx= 1 a tan 1 x a 1 (10) z x a2 + x2 dx= 1 2 lnja2 + x2j (11) z. Z tanxdx= ln cosx +c 7. Table of basic integrals basic forms (1) z xndx= 1 n+ 1 xn+1; Sn+1 (11) tx (x 1 2r) ( x+ 1) sx+1 (12) sinkt k s2 + k2 (13) coskt s s2 + k2 (14) eat 1 s a (15) sinhkt k. Table 2.1, choose yp in the same line and determine its undetermined coefficients by substituting yp and its derivatives into (4). Z dx x = lnjxj+c 3. Udv a b ∫=#uv$% a b −vdu a b ∫ u and v are. Csun, integrals, table of integrals, math 280, math 351, differential equations created date: N 6= 1 (2) z 1 x dx = lnjxj (3) z u dv = uv z vdu (4) z e xdx = e (5) z ax dx = 1 lna ax (6) z lnxdx = xlnx x (7) z sinxdx = cosx (8) z cosxdx = sinx (9) z tanxdx = lnjsecxj (10) z secxdx = lnjsecx+tanxj (11) z sec2 xdx = tanx (12) z secxtanxdx = secx (13) z a a2 +x2 dx = tan 1 x a (14) z a a2 x2 dx = 1 2 ln x+a x a (15) z 1 p a2 2x dx = sin. E−ax2dx= 1 2 π a # $% & '(1 2 0 ∞ ∫ ax xe−2dx= 1 2a 0 ∞ ∫ x2e−ax2dx= 1 4a π a # $% & '(1 2 0 ∞ ∫ x3e−ax2dx= 1 2a2 0 ∞ ∫ x2ne−ax2dx= 1⋅3⋅5⋅⋅⋅(2n−1) 2n+1an π a $ %& ' 1 2 0 ∞ ∫ x2n+1e−ax2dx= n! Table of integrals basic forms z xndx= 1 n+ 1 xn+1 (1) z 1 x dx= lnx (2) z udv= uv z vdu (3) z 1 ax+ b dx= 1 a lnjax+ bj (4) integrals of rational functions z 1 (x+ a)2 dx= 1 x+ a (5) z (x+ a)ndx= (x+ a)n+1 n+ 1 + c;n6= 1 (6) z x(x+ a)ndx= (x+ a)n+1((n+ 1)x a) (n+ 1)(n+ 2) (7) z 1 1 + x2 dx= tan 1 x (8) z 1 a2 + x2 dx= 1 a tan 1 x a (9) z x a 2+ x dx= 1 2 lnja2 + x2j (10) z x 2 a 2+ x dx= x.
Z secxdx= ln secx+tanx +c 12. Sn+1 (11) tx (x 1 2r) ( x+ 1) sx+1 (12) sinkt k s2 + k2 (13) coskt s s2 + k2 (14) eat 1 s a (15) sinhkt k. Z dx a 2+x = 1 a tan 1 x a +c 9. The copyright holder makes no representation about the accuracy, correctness, or 2an+1 0 ∞ ∫ xne−axdx= n!
N6= 1 (2) z 1 x dx= lnjxj (3) z udv= uv z vdu (4) z 1 ax+ b dx= 1 a lnjax+ bj integrals of rational functions (5) z 1 (x+ a)2 dx= 1 x+ a (6) z (x+ a)ndx= (x+ a)n+1 n+ 1;n6= 1 (7) z x(x+ a)ndx= (x+ a)n+1((n+ 1)x a) (n+ 1)(n+ 2) (8) z 1 1 + x2 dx= tan 1 x (9) z 1 a2 + x2 dx= 1 a tan 1 x a 1 (10) z x a2 + x2 dx= 1 2 lnja2 + x2j (11) z.
Z xn dx= xn+1 n+1 +c (n6= 1) 2. 2an+1 0 ∞ ∫ xne−axdx= n! C) irrationale funktionen 22) ax = b dx (ax b) 3/2 3a 2 23) dx ax b 1 = ax b a 2 24) x = ax b dx 3/2 2 (3 ax 2 b) (ax b) 15 a 2 25) dx ax b x = (a x 2 b) ax b 3a 2 2 26) dx x ax b 1 = ax b b ax b b ln b 1 für b > 0 = b ax b arctan b 2 für b < 0 27) ax b dx x 1 = dx x ax b 1 2. Z dx x = lnjxj+c 3. Z tanxdx= ln cosx +c 7. N 6= 1 (2) z 1 x dx = lnjxj (3) z u dv = uv z vdu (4) z e xdx = e (5) z ax dx = 1 lna ax (6) z lnxdx = xlnx x (7) z sinxdx = cosx (8) z cosxdx = sinx (9) z tanxdx = lnjsecxj (10) z secxdx = lnjsecx+tanxj (11) z sec2 xdx = tanx (12) z secxtanxdx = secx (13) z a a2 +x2 dx = tan 1 x a (14) z a a2 x2 dx = 1 2 ln x+a x a (15) z 1 p a2 2x dx = sin. 01.07.2016 · table of laplace transforms f(t) lf(t) = f(s) 1 1 s (1) eatf(t) f(s a) (2) u(t a) e as s (3) f(t a)u(t a) e asf(s) (4) (t) 1 (5) (t stt 0) e 0 (6) tnf(t) ( 1)n dnf(s) dsn (7) f0(t) sf(s) f(0) (8) fn(t) snf(s) s(n 1)f(0) (fn 1)(0) (9) z t 0 f(x)g(t x)dx f(s)g(s) (10) tn (n= 0;1;2;:::) n! Table 2.1, choose yp in the same line and determine its undetermined coefficients by substituting yp and its derivatives into (4). The copyright holder makes no representation about the accuracy, correctness, or Csun, integrals, table of integrals, math 280, math 351, differential equations created date: Z cosecxdx= ln cosecx cotx +c 13. Table of basic integrals1 (1) z xn dx = 1 n+1 xn+1; Sn+1 (11) tx (x 1 2r) ( x+ 1) sx+1 (12) sinkt k s2 + k2 (13) coskt s s2 + k2 (14) eat 1 s a (15) sinhkt k.
Bei integralen über echt gebrochenrationale funktionen wird auf die methode der partialbruchzerlegung verwiesen. E−ax2dx= 1 2 π a # $% & '(1 2 0 ∞ ∫ ax xe−2dx= 1 2a 0 ∞ ∫ x2e−ax2dx= 1 4a π a # $% & '(1 2 0 ∞ ∫ x3e−ax2dx= 1 2a2 0 ∞ ∫ x2ne−ax2dx= 1⋅3⋅5⋅⋅⋅(2n−1) 2n+1an π a $ %& ' 1 2 0 ∞ ∫ x2n+1e−ax2dx= n! Udv a b ∫=#uv$% a b −vdu a b ∫ u and v are. Csun, integrals, table of integrals, math 280, math 351, differential equations created date: Table of basic integrals basic forms (1) z xndx= 1 n+ 1 xn+1;
Z secxdx= ln secx+tanx +c 12.
N 6= 1 (2) z 1 x dx = lnjxj (3) z u dv = uv z vdu (4) z e xdx = e (5) z ax dx = 1 lna ax (6) z lnxdx = xlnx x (7) z sinxdx = cosx (8) z cosxdx = sinx (9) z tanxdx = lnjsecxj (10) z secxdx = lnjsecx+tanxj (11) z sec2 xdx = tanx (12) z secxtanxdx = secx (13) z a a2 +x2 dx = tan 1 x a (14) z a a2 x2 dx = 1 2 ln x+a x a (15) z 1 p a2 2x dx = sin. Z secxdx= ln secx+tanx +c 12. Table of basic integrals1 (1) z xn dx = 1 n+1 xn+1; Z cosecxdx= ln cosecx cotx +c 13. Bei integralen über echt gebrochenrationale funktionen wird auf die methode der partialbruchzerlegung verwiesen. An+1 0 ∞ ∫ integration by parts: The copyright holder makes no representation about the accuracy, correctness, or Table 2.1, choose yp in the same line and determine its undetermined coefficients by substituting yp and its derivatives into (4). N6= 1 (2) z 1 x dx= lnjxj (3) z udv= uv z vdu (4) z 1 ax+ b dx= 1 a lnjax+ bj integrals of rational functions (5) z 1 (x+ a)2 dx= 1 x+ a (6) z (x+ a)ndx= (x+ a)n+1 n+ 1;n6= 1 (7) z x(x+ a)ndx= (x+ a)n+1((n+ 1)x a) (n+ 1)(n+ 2) (8) z 1 1 + x2 dx= tan 1 x (9) z 1 a2 + x2 dx= 1 a tan 1 x a 1 (10) z x a2 + x2 dx= 1 2 lnja2 + x2j (11) z. Sn+1 (11) tx (x 1 2r) ( x+ 1) sx+1 (12) sinkt k s2 + k2 (13) coskt s s2 + k2 (14) eat 1 s a (15) sinhkt k. Table of basic integrals basic forms (1) z xndx= 1 n+ 1 xn+1; Z e xdx= e +c 4. Z cotxdx= ln sinx +c 8.
Integral Table Pdf : Ejercicios propuestos de Integrales, Método de Sustitución / ©2005 be shapiro page 3 this document may not be reproduced, posted or published without permission.. Z xn dx= xn+1 n+1 +c (n6= 1) 2. Z dx x = lnjxj+c 3. N6= 1 (2) z 1 x dx= lnjxj (3) z udv= uv z vdu (4) z 1 ax+ b dx= 1 a lnjax+ bj integrals of rational functions (5) z 1 (x+ a)2 dx= 1 x+ a (6) z (x+ a)ndx= (x+ a)n+1 n+ 1;n6= 1 (7) z x(x+ a)ndx= (x+ a)n+1((n+ 1)x a) (n+ 1)(n+ 2) (8) z 1 1 + x2 dx= tan 1 x (9) z 1 a2 + x2 dx= 1 a tan 1 x a 1 (10) z x a2 + x2 dx= 1 2 lnja2 + x2j (11) z. Csun, integrals, table of integrals, math 280, math 351, differential equations created date: Bei integralen über echt gebrochenrationale funktionen wird auf die methode der partialbruchzerlegung verwiesen.
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