Theoretical Computer Science Cheat Sheet, Cheat Sheet of Computer Science

Download Theoretical Computer Science Cheat Sheet and more Cheat Sheet Computer Science in PDF only on Docsity! Theoretical Computer Science Cheat Sheet Definitions Series f(n) = O(g(n)) iff ∃ positive c, n0 such that 0 ≤ f(n) ≤ cg(n) ∀n ≥ n0. n ∑ i=1 i = n(n+ 1) 2 , n ∑ i=1 i2 = n(n+ 1)(2n+ 1) 6 , n ∑ i=1 i3 = n2(n+ 1)2 4 . In general: n ∑ i=1 im = 1 m+ 1 [ (n+ 1)m+1 − 1− n ∑ i=1 ( (i+ 1)m+1 − im+1 − (m+ 1)im ) ] n−1 ∑ i=1 im = 1 m+ 1 m ∑ k=0 ( m+ 1 k ) Bkn m+1−k. Geometric series: n ∑ i=0 ci = cn+1 − 1 c− 1 , c 6= 1, ∞ ∑ i=0 ci = 1 1− c , ∞ ∑ i=1 ci = c 1− c , |c| < 1, n ∑ i=0 ici = ncn+2 − (n+ 1)cn+1 + c (c− 1)2 , c 6= 1, ∞ ∑ i=0 ici = c (1− c)2 , |c| < 1. Harmonic series: Hn = n ∑ i=1 1 i , n ∑ i=1 iHi = n(n+ 1) 2 Hn − n(n− 1) 4 . n ∑ i=1 Hi = (n+ 1)Hn − n, n ∑ i=1 ( i m ) Hi = ( n+ 1 m+ 1 )( Hn+1 − 1 m+ 1 ) . f(n) = Ω(g(n)) iff ∃ positive c, n0 such that f(n) ≥ cg(n) ≥ 0 ∀n ≥ n0. f(n) = Θ(g(n)) iff f(n) = O(g(n)) and f(n) = Ω(g(n)). f(n) = o(g(n)) iff limn→∞ f(n)/g(n) = 0. lim n→∞ an = a iff ∀ǫ > 0, ∃n0 such that |an − a| < ǫ, ∀n ≥ n0. supS least b ∈ R such that b ≥ s, ∀s ∈ S. inf S greatest b ∈ R such that b ≤ s, ∀s ∈ S. lim inf n→∞ an lim n→∞ inf{ai | i ≥ n, i ∈ N}. lim sup n→∞ an lim n→∞ sup{ai | i ≥ n, i ∈ N}. ( n k ) Combinations: Size k sub- sets of a size n set. [ n k ] Stirling numbers (1st kind): Arrangements of an n ele- ment set into k cycles. 1. ( n k ) = n! (n− k)!k! , 2. n ∑ k=0 ( n k ) = 2n, 3. ( n k ) = ( n n− k ) , 4. ( n k ) = n k ( n− 1 k − 1 ) , 5. ( n k ) = ( n− 1 k ) + ( n− 1 k − 1 ) , 6. ( n m )( m k ) = ( n k )( n− k m− k ) , 7. n ∑ k=0 ( r + k k ) = ( r + n+ 1 n ) , 8. n ∑ k=0 ( k m ) = ( n+ 1 m+ 1 ) , 9. n ∑ k=0 ( r k )( s n− k ) = ( r + s n ) , 10. ( n k ) = (−1)k ( k − n− 1 k ) , 11. { n 1 } = { n n } = 1, 12. { n 2 } = 2n−1 − 1, 13. { n k } = k { n− 1 k } + { n− 1 k − 1 } , { n k } Stirling numbers (2nd kind): Partitions of an n element set into k non-empty sets. 〈 n k 〉 1st order Eulerian numbers: Permutations π1π2 . . . πn on {1, 2, . . . , n} with k ascents. 〈 n k 〉 2nd order Eulerian numbers. Cn Catalan Numbers: Binary trees with n+ 1 vertices. 14. [ n 1 ] = (n− 1)!, 15. [ n 2 ] = (n− 1)!Hn−1, 16. [ n n ] = 1, 17. [ n k ] ≥ { n k } , 18. [ n k ] = (n− 1) [ n− 1 k ] + [ n− 1 k − 1 ] , 19. { n n− 1 } = [ n n− 1 ] = ( n 2 ) , 20. n ∑ k=0 [ n k ] = n!, 21. Cn = 1 n+ 1 ( 2n n ) , 22. 〈 n 0 〉 = 〈 n n− 1 〉 = 1, 23. 〈 n k 〉 = 〈 n n− 1− k 〉 , 24. 〈 n k 〉 = (k + 1) 〈 n− 1 k 〉 + (n− k) 〈 n− 1 k − 1 〉 , 25. 〈 0 k 〉 = { 1 if k = 0, 0 otherwise 26. 〈 n 1 〉 = 2n − n− 1, 27. 〈 n 2 〉 = 3n − (n+ 1)2n + ( n+ 1 2 ) , 28. xn = n ∑ k=0 〈 n k 〉( x+ k n ) , 29. 〈 n m 〉 = m ∑ k=0 ( n+ 1 k ) (m+ 1− k)n(−1)k, 30. m! { n m } = n ∑ k=0 〈 n k 〉( k n−m ) , 31. 〈 n m 〉 = n ∑ k=0 { n k }( n− k m ) (−1)n−k−mk!, 32. 〈〈 n 0 〉〉 = 1, 33. 〈〈 n n 〉〉 = 0 for n 6= 0, 34. 〈〈 n k 〉〉 = (k + 1) 〈〈 n− 1 k 〉 + (2n− 1− k) 〈〈 n− 1 k − 1 〉〉 , 35. n ∑ k=0 〈〈 n k 〉〉 = (2n)n 2n , 36. { x x− n } = n ∑ k=0 〈 n k 〉〉( x+ n− 1− k 2n ) , 37. { n+ 1 m+ 1 } = ∑ k ( n k ){ k m } = n ∑ k=0 { k m } (m+ 1)n−k, Theoretical Computer Science Cheat Sheet Identities Cont. Trees 38. [ n+ 1 m+ 1 ] = ∑ k [ n k ]( k m ) = n ∑ k=0 [ k m ] nn−k = n! n ∑ k=0 1 k! [ k m ] , 39. [ x x− n ] = n ∑ k=0 〈〈 n k 〉〉( x+ k 2n ) , 40. { n m } = ∑ k ( n k ){ k + 1 m+ 1 } (−1)n−k, 41. [ n m ] = ∑ k [ n+ 1 k + 1 ]( k m ) (−1)m−k, 42. { m+ n+ 1 m } = m ∑ k=0 k { n+ k k } , 43. [ m+ n+ 1 m ] = m ∑ k=0 k(n+ k) [ n+ k k ] , 44. ( n m ) = ∑ k { n+ 1 k + 1 }[ k m ] (−1)m−k, 45. (n−m)! ( n m ) = ∑ k [ n+ 1 k + 1 ]{ k m } (−1)m−k, for n ≥ m, 46. { n n−m } = ∑ k ( m− n m+ k )( m+ n n+ k )[ m+ k k ] , 47. [ n n−m ] = ∑ k ( m− n m+ k )( m+ n n+ k ){ m+ k k } , 48. { n ℓ+m }( ℓ+m ℓ ) = ∑ k { k ℓ }{ n− k m }( n k ) , 49. [ n ℓ+m ]( ℓ+m ℓ ) = ∑ k [ k ℓ ][ n− k m ]( n k ) . Every tree with n vertices has n − 1 edges. Kraft inequal- ity: If the depths of the leaves of a binary tree are d1, . . . , dn: n ∑ i=1 2−di ≤ 1, and equality holds only if every in- ternal node has 2 sons. Recurrences Master method: T (n) = aT (n/b) + f(n), a ≥ 1, b > 1 If ∃ǫ > 0 such that f(n) = O(nlogb a−ǫ) then T (n) = Θ(nlogb a). If f(n) = Θ(nlogb a) then T (n) = Θ(nlogb a log2 n). If ∃ǫ > 0 such that f(n) = Ω(nlogb a+ǫ), and ∃c < 1 such that af(n/b) ≤ cf(n) for large n, then T (n) = Θ(f(n)). Substitution (example): Consider the following recurrence Ti+1 = 2 2i · T 2i , T1 = 2. Note that Ti is always a power of two. Let ti = log2 Ti. Then we have ti+1 = 2 i + 2ti, t1 = 1. Let ui = ti/2 i. Dividing both sides of the previous equation by 2i+1 we get ti+1 2i+1 = 2i 2i+1 + ti 2i . Substituting we find ui+1 = 1 2 + ui, u1 = 1 2 , which is simply ui = i/2. So we find that Ti has the closed form Ti = 2 i2i−1 . Summing factors (example): Consider the following recurrence T (n) = 3T (n/2) + n, T (1) = 1. Rewrite so that all terms involving T are on the left side T (n)− 3T (n/2) = n. Now expand the recurrence, and choose a factor which makes the left side “tele- scope” 1 ( T (n)− 3T (n/2) = n ) 3 ( T (n/2)− 3T (n/4) = n/2 ) ... ... ... 3log2 n−1 ( T (2)− 3T (1) = 2 ) Let m = log2 n. Summing the left side we get T (n) − 3mT (1) = T (n) − 3m = T (n) − nk where k = log2 3 ≈ 1.58496. Summing the right side we get m−1 ∑ i=0 n 2i 3i = n m−1 ∑ i=0 ( 3 2 )i . Let c = 32 . Then we have n m−1 ∑ i=0 ci = n ( cm − 1 c− 1 ) = 2n(clog2 n − 1) = 2n(c(k−1) logc n − 1) = 2nk − 2n, and so T (n) = 3nk − 2n. Full history re- currences can often be changed to limited history ones (example): Consider Ti = 1 + i−1 ∑ j=0 Tj, T0 = 1. Note that Ti+1 = 1 + i ∑ j=0 Tj. Subtracting we find Ti+1 − Ti = 1 + i ∑ j=0 Tj − 1− i−1 ∑ j=0 Tj = Ti. And so Ti+1 = 2Ti = 2 i+1. Generating functions: 1. Multiply both sides of the equa- tion by xi. 2. Sum both sides over all i for which the equation is valid. 3. Choose a generating function G(x). Usually G(x) = ∑∞ i=0 x igi. 3. Rewrite the equation in terms of the generating function G(x). 4. Solve for G(x). 5. The coefficient of xi in G(x) is gi. Example: gi+1 = 2gi + 1, g0 = 0. Multiply and sum: ∑ i≥0 gi+1x i = ∑ i≥0 2gix i + ∑ i≥0 xi. We chooseG(x) = ∑ i≥0 x igi. Rewrite in terms of G(x): G(x) − g0 x = 2G(x) + ∑ i≥0 xi. Simplify: G(x) x = 2G(x) + 1 1− x . Solve for G(x): G(x) = x (1− x)(1 − 2x) . Expand this using partial fractions: G(x) = x ( 2 1− 2x − 1 1− x ) = x  2 ∑ i≥0 2ixi − ∑ i≥0 xi   = ∑ i≥0 (2i+1 − 1)xi+1. So gi = 2 i − 1. Theoretical Computer Science Cheat Sheet Number Theory Graph Theory The Chinese remainder theorem: There ex- ists a number C such that: C ≡ r1 mod m1 ... ... ... C ≡ rn mod mn if mi and mj are relatively prime for i 6= j. Euler’s function: φ(x) is the number of positive integers less than x relatively prime to x. If ∏n i=1 p ei i is the prime fac- torization of x then φ(x) = n ∏ i=1 pei−1i (pi − 1). Euler’s theorem: If a and b are relatively prime then 1 ≡ aφ(b) mod b. Fermat’s theorem: 1 ≡ ap−1 mod p. The Euclidean algorithm: if a > b are in- tegers then gcd(a, b) = gcd(a mod b, b). If ∏n i=1 p ei i is the prime factorization of x then S(x) = ∑ d|x d = n ∏ i=1 pei+1i − 1 pi − 1 . Perfect Numbers: x is an even perfect num- ber iff x = 2n−1(2n−1) and 2n−1 is prime. Wilson’s theorem: n is a prime iff (n− 1)! ≡ −1 mod n. Möbius inversion: µ(i) =      1 if i = 1. 0 if i is not square-free. (−1)r if i is the product of r distinct primes. If G(a) = ∑ d|a F (d), then F (a) = ∑ d|a µ(d)G (a d ) . Prime numbers: pn = n lnn+ n ln lnn− n+ n ln lnn lnn +O ( n lnn ) , π(n) = n lnn + n (lnn)2 + 2!n (lnn)3 +O ( n (lnn)4 ) . Definitions: Loop An edge connecting a ver- tex to itself. Directed Each edge has a direction. Simple Graph with no loops or multi-edges. Walk A sequence v0e1v1 . . . eℓvℓ. Trail A walk with distinct edges. Path A trail with distinct vertices. Connected A graph where there exists a path between any two vertices. Component A maximal connected subgraph. Tree A connected acyclic graph. Free tree A tree with no root. DAG Directed acyclic graph. Eulerian Graph with a trail visiting each edge exactly once. Hamiltonian Graph with a cycle visiting each vertex exactly once. Cut A set of edges whose re- moval increases the num- ber of components. Cut-set A minimal cut. Cut edge A size 1 cut. k-Connected A graph connected with the removal of any k − 1 vertices. k-Tough ∀S ⊆ V, S 6= ∅ we have k · c(G− S) ≤ |S|. k-Regular A graph where all vertices have degree k. k-Factor A k-regular spanning subgraph. Matching A set of edges, no two of which are adjacent. Clique A set of vertices, all of which are adjacent. Ind. set A set of vertices, none of which are adjacent. Vertex cover A set of vertices which cover all edges. Planar graph A graph which can be em- beded in the plane. Plane graph An embedding of a planar graph. ∑ v∈V deg(v) = 2m. If G is planar then n−m+ f = 2, so f ≤ 2n− 4, m ≤ 3n− 6. Any planar graph has a vertex with de- gree ≤ 5. Notation: E(G) Edge set V (G) Vertex set c(G) Number of components G[S] Induced subgraph deg(v) Degree of v ∆(G) Maximum degree δ(G) Minimum degree χ(G) Chromatic number χE(G) Edge chromatic number Gc Complement graph Kn Complete graph Kn1,n2 Complete bipartite graph r(k, ℓ) Ramsey number Geometry Projective coordinates: triples (x, y, z), not all x, y and z zero. (x, y, z) = (cx, cy, cz) ∀c 6= 0. Cartesian Projective (x, y) (x, y, 1) y = mx+ b (m,−1, b) x = c (1, 0,−c) Distance formula, Lp and L∞ metric: √ (x1 − x0)2 + (y1 − y0)2, [ |x1 − x0|p + |y1 − y0|p ]1/p , lim p→∞ [ |x1 − x0|p + |y1 − y0|p ]1/p . Area of triangle (x0, y0), (x1, y1) and (x2, y2): 1 2 abs ∣ ∣ ∣ ∣ x1 − x0 y1 − y0 x2 − x0 y2 − y0 ∣ ∣ ∣ ∣ . Angle formed by three points: (0, 0) θ (x1, y1) (x2, y2) ℓ2 ℓ1 cos θ = (x1, y1) · (x2, y2) ℓ1ℓ2 . Line through two points (x0, y0) and (x1, y1): ∣ ∣ ∣ ∣ ∣ ∣ x y 1 x0 y0 1 x1 y1 1 ∣ ∣ ∣ ∣ ∣ ∣ = 0. Area of circle, volume of sphere: A = πr2, V = 43πr 3. If I have seen farther than others, it is because I have stood on the shoulders of giants. – Issac Newton Theoretical Computer Science Cheat Sheet π Calculus Wallis’ identity: π = 2 · 2 · 2 · 4 · 4 · 6 · 6 · · · 1 · 3 · 3 · 5 · 5 · 7 · · · Brouncker’s continued fraction expansion: π 4 = 1 + 12 2 + 3 2 2+ 5 2 2+ 7 2 2+··· Gregrory’s series: π 4 = 1− 13 + 15 − 17 + 19 − · · · Newton’s series: π 6 = 1 2 + 1 2 · 3 · 23 + 1 · 3 2 · 4 · 5 · 25 + · · · Sharp’s series: π 6 = 1√ 3 ( 1− 1 31 · 3 + 1 32 · 5 − 1 33 · 7 + · · · ) Euler’s series: π2 6 = 1 12 + 1 22 + 1 32 + 1 42 + 1 52 + · · · π2 8 = 1 12 + 1 32 + 1 52 + 1 72 + 1 92 + · · · π2 12 = 1 12 − 122 + 132 − 142 + 152 − · · · Derivatives: 1. d(cu) dx = c du dx , 2. d(u + v) dx = du dx + dv dx , 3. d(uv) dx = u dv dx + v du dx , 4. d(un) dx = nun−1 du dx , 5. d(u/v) dx = v ( du dx ) − u ( dv dx ) v2 , 6. d(ecu) dx = cecu du dx , 7. d(cu) dx = (ln c)cu du dx , 8. d(ln u) dx = 1 u du dx , 9. d(sinu) dx = cosu du dx , 10. d(cosu) dx = − sinudu dx , 11. d(tanu) dx = sec2 u du dx , 12. d(cotu) dx = csc2 u du dx , 13. d(secu) dx = tanu secu du dx , 14. d(cscu) dx = − cotu cscudu dx , 15. d(arcsinu) dx = 1√ 1− u2 du dx , 16. d(arccosu) dx = −1√ 1− u2 du dx , 17. d(arctanu) dx = 1 1 + u2 du dx , 18. d(arccotu) dx = −1 1 + u2 du dx , 19. d(arcsecu) dx = 1 u √ 1− u2 du dx , 20. d(arccscu) dx = −1 u √ 1− u2 du dx , 21. d(sinhu) dx = coshu du dx , 22. d(coshu) dx = sinhu du dx , 23. d(tanhu) dx = sech2 u du dx , 24. d(cothu) dx = − csch2 udu dx , 25. d(sechu) dx = − sechu tanhudu dx , 26. d(cschu) dx = − cschu cothudu dx , 27. d(arcsinhu) dx = 1√ 1 + u2 du dx , 28. d(arccoshu) dx = 1√ u2 − 1 du dx , 29. d(arctanhu) dx = 1 1− u2 du dx , 30. d(arccothu) dx = 1 u2 − 1 du dx , 31. d(arcsechu) dx = −1 u √ 1− u2 du dx , 32. d(arccschu) dx = −1 |u| √ 1 + u2 du dx . Integrals: 1. ∫ cu dx = c ∫ u dx, 2. ∫ (u + v) dx = ∫ u dx+ ∫ v dx, 3. ∫ xn dx = 1 n+ 1 xn+1, n 6= −1, 4. ∫ 1 x dx = lnx, 5. ∫ ex dx = ex, 6. ∫ dx 1 + x2 = arctanx, 7. ∫ u dv dx dx = uv − ∫ v du dx dx, 8. ∫ sinx dx = − cosx, 9. ∫ cosx dx = sinx, 10. ∫ tanx dx = − ln | cosx|, 11. ∫ cotx dx = ln | cosx|, 12. ∫ secx dx = ln | secx+ tanx|, 13. ∫ cscx dx = ln | cscx+ cotx|, 14. ∫ arcsin xadx = arcsin x a + √ a2 − x2, a > 0, Partial Fractions Let N(x) and D(x) be polynomial func- tions of x. We can break down N(x)/D(x) using partial fraction expan- sion. First, if the degree of N is greater than or equal to the degree of D, divide N by D, obtaining N(x) D(x) = Q(x) + N ′(x) D(x) , where the degree of N ′ is less than that of D. Second, factor D(x). Use the follow- ing rules: For a non-repeated factor: N(x) (x− a)D(x) = A x− a + N ′(x) D(x) , where A = [ N(x) D(x) ] x=a . For a repeated factor: N(x) (x− a)mD(x) = m−1 ∑ k=0 Ak (x− a)m−k+ N ′(x) D(x) , where Ak = 1 k! [ dk dxk ( N(x) D(x) )] x=a . The reasonable man adapts himself to the world; the unreasonable persists in trying to adapt the world to himself. Therefore all progress depends on the unreasonable. – George Bernard Shaw Theoretical Computer Science Cheat Sheet Calculus Cont. 15. ∫ arccos xadx = arccos x a − √ a2 − x2, a > 0, 16. ∫ arctan xadx = x arctan x a − a2 ln(a2 + x2), a > 0, 17. ∫ sin2(ax)dx = 12a ( ax− sin(ax) cos(ax) ) , 18. ∫ cos2(ax)dx = 12a ( ax+ sin(ax) cos(ax) ) , 19. ∫ sec2 x dx = tanx, 20. ∫ csc2 x dx = − cotx, 21. ∫ sinn x dx = − sin n−1 x cosx n + n− 1 n ∫ sinn−2 x dx, 22. ∫ cosn x dx = cosn−1 x sinx n + n− 1 n ∫ cosn−2 x dx, 23. ∫ tann x dx = tann−1 x n− 1 − ∫ tann−2 x dx, n 6= 1, 24. ∫ cotn x dx = −cot n−1 x n− 1 − ∫ cotn−2 x dx, n 6= 1, 25. ∫ secn x dx = tanx secn−1 x n− 1 + n− 2 n− 1 ∫ secn−2 x dx, n 6= 1, 26. ∫ cscn x dx = −cotx csc n−1 x n− 1 + n− 2 n− 1 ∫ cscn−2 x dx, n 6= 1, 27. ∫ sinhx dx = coshx, 28. ∫ coshx dx = sinhx, 29. ∫ tanhx dx = ln | coshx|, 30. ∫ cothx dx = ln | sinhx|, 31. ∫ sechx dx = arctan sinhx, 32. ∫ cschx dx = ln ∣ ∣tanh x2 ∣ ∣ , 33. ∫ sinh2 x dx = 14 sinh(2x)− 12x, 34. ∫ cosh2 x dx = 14 sinh(2x) + 1 2x, 35. ∫ sech2 x dx = tanhx, 36. ∫ arcsinh xadx = x arcsinh x a − √ x2 + a2, a > 0, 37. ∫ arctanh xadx = x arctanh x a + a 2 ln |a2 − x2|, 38. ∫ arccosh xadx =    x arccosh x a − √ x2 + a2, if arccosh xa > 0 and a > 0, x arccosh x a + √ x2 + a2, if arccosh xa < 0 and a > 0, 39. ∫ dx√ a2 + x2 = ln ( x+ √ a2 + x2 ) , a > 0, 40. ∫ dx a2 + x2 = 1a arctan x a , a > 0, 41. ∫ √ a2 − x2 dx = x2 √ a2 − x2 + a22 arcsin xa , a > 0, 42. ∫ (a2 − x2)3/2dx = x8 (5a2 − 2x2) √ a2 − x2 + 3a48 arcsin xa , a > 0, 43. ∫ dx√ a2 − x2 = arcsin xa , a > 0, 44. ∫ dx a2 − x2 = 1 2a ln ∣ ∣ ∣ ∣ a+ x a− x ∣ ∣ ∣ ∣ , 45. ∫ dx (a2 − x2)3/2 = x a2 √ a2 − x2 , 46. ∫ √ a2 ± x2 dx = x2 √ a2 ± x2 ± a22 ln ∣ ∣ ∣x+ √ a2 ± x2 ∣ ∣ ∣ , 47. ∫ dx√ x2 − a2 = ln ∣ ∣ ∣x+ √ x2 − a2 ∣ ∣ ∣ , a > 0, 48. ∫ dx ax2 + bx = 1 a ln ∣ ∣ ∣ ∣ x a+ bx ∣ ∣ ∣ ∣ , 49. ∫ x √ a+ bx dx = 2(3bx− 2a)(a+ bx)3/2 15b2 , 50. ∫ √ a+ bx x dx = 2 √ a+ bx+ a ∫ 1 x √ a+ bx dx, 51. ∫ x√ a+ bx dx = 1√ 2 ln ∣ ∣ ∣ ∣ √ a+ bx−√a√ a+ bx+ √ a ∣ ∣ ∣ ∣ , a > 0, 52. ∫ √ a2 − x2 x dx = √ a2 − x2 − a ln ∣ ∣ ∣ ∣ ∣ a+ √ a2 − x2 x ∣ ∣ ∣ ∣ ∣ , 53. ∫ x √ a2 − x2 dx = − 13 (a2 − x2)3/2, 54. ∫ x2 √ a2 − x2 dx = x8 (2x2 − a2) √ a2 − x2 + a48 arcsin xa , a > 0, 55. ∫ dx√ a2 − x2 = − 1a ln ∣ ∣ ∣ ∣ ∣ a+ √ a2 − x2 x ∣ ∣ ∣ ∣ ∣ , 56. ∫ x dx√ a2 − x2 = − √ a2 − x2, 57. ∫ x2 dx√ a2 − x2 = −x2 √ a2 − x2 + a22 arcsin xa, a > 0, 58. ∫ √ a2 + x2 x dx = √ a2 + x2 − a ln ∣ ∣ ∣ ∣ ∣ a+ √ a2 + x2 x ∣ ∣ ∣ ∣ ∣ , 59. ∫ √ x2 − a2 x dx = √ x2 − a2 − a arccos a|x| , a > 0, 60. ∫ x √ x2 ± a2 dx = 13 (x2 ± a2)3/2, 61. ∫ dx x √ x2 + a2 = 1a ln ∣ ∣ ∣ ∣ x a+ √ a2 + x2 ∣ ∣ ∣ ∣ ,