Determine la integral indefinida 𝑛 1) ∫ 𝑢 𝑑𝑢 = 𝑢𝑛+1 ; 2) ∫ 𝑛+1 𝑑𝑢 = 𝑙𝑛|𝑢|;3) ∫ 𝑢 𝑑𝑢 𝑎2 +𝑢2 = 𝑎𝑟𝑐𝑡𝑎𝑛 𝑢 𝑎 53) ∫(2𝑡𝑖 + 𝑗 + 𝑘)𝑑𝑡=∫ 2𝑡𝑖𝑑𝑡+∫ 𝑗𝑑𝑡+∫ 𝑘𝑑𝑡 2𝑡 2 2 i+tj+tk+C=𝑡 2 𝑖+tj+tk+C 3 2 1 55) ∫ ( 𝑖 + 𝑗 − 𝑡 𝑘) 𝑑𝑡=∫ 𝑡 = 𝑙𝑛|𝑢|i+tj- 𝑑𝑡 𝑡 3 2 𝑖+∫ 𝑗𝑑𝑡 − ∫ 𝑡 𝑘𝑑𝑡 5 5 𝑡2 2𝑡 2 3 2 𝑘= 𝑙𝑛|𝑢|i+tj - 3 𝑘+C 57)∫[(2𝑡 − 1)𝑖 + 4𝑡 3 𝑗 + 3√𝑡𝑘]dt (∫ 2𝑡 𝑑𝑡 − ∫ 𝑑𝑡)𝑖 + ∫ 4𝑡 3 𝑗 + ∫ 3√𝑡𝑘𝑑𝑡= 4𝑡 4 (𝑡 2 − 𝑡)𝑖+ 4 3 𝑗+ 59 ∫ (𝑠𝑒𝑐 2 𝑡𝑖 + 3𝑡 2 3 2 k+C=(𝑡 2 − 𝑡)𝑖+𝑡 4 𝑗+2√𝑡 3 𝑘+C 1 1+𝑡 2 𝑗) 𝑑𝑡=∫ 𝑠𝑒𝑐 2 𝑡𝑖𝑑𝑡 + ∫ 𝑑𝑡 1+𝑡 2 𝑗 =tanti+arctan𝑡 2 𝑗+C 𝜋 2 𝜋 2 ∫0 [(𝑎𝑐𝑜𝑠𝑡)𝑖 + (𝑎𝑠𝑒𝑛𝑡)𝑗 + 𝑘]𝑑𝑡 = [𝑎𝑠𝑒𝑛𝑡𝑖 − 𝑎𝑐𝑜𝑠𝑡𝑗 + 𝑡𝑘 ]0 = 𝜋 𝜋 𝜋 2 2 2 [𝑎𝑠𝑒𝑛 𝑖 − 𝑎𝑐𝑜𝑠 𝑗 + 𝑘] − [𝑎𝑠𝑒𝑛0𝑖 − 𝑎𝑐𝑜𝑠0𝑗 + 𝑜𝑘] 𝜋 𝜋 2 2 [ai-0+ 𝑘] − [0 − 𝑎𝑗 + 0=ai-aj+ 𝑘 2 𝑥 ∫1 [∫1 (2𝑥 2 𝑦 −2 + 2𝑦)𝑑𝑦]𝑑𝑥 = 𝑥 ∫1 (2𝑥 2 𝑦 −2 + 2𝑦)𝑑𝑦=[ 2𝑥 2 𝑦 −1 −1 + 2𝑦 2 2 𝑥 ] =[ 1 2𝑥 2 −𝑦 2 𝑥 +𝑦 ] [−2𝑥 + 𝑥 2 ] − [−2𝑥 2 + 1] =−2𝑥 + 𝑥 2 +2𝑥 2 − 1 1 =3𝑥 2 -2x2 ∫ [3𝑥 2 − 2x − 1]𝑑𝑥 = [𝑥 3 − 𝑥 2 − 𝑥 ]12 1 =[(2)3 − (2)2 − 2]- [] − [1-1-1]==2+11=3 𝑥 ∫0 (𝑥 + 2𝑦)𝑑𝑦= [𝑥𝑦 + 𝑦 2 ]0𝑥 =𝑥 2 + 𝑥 2 − [0 + 0]=2𝑥 2 2𝑦 𝑦 ∫1 𝑥 1 2 2𝑦 𝑑𝑥 =[𝑦𝑙𝑛|𝑥 |]1 =y 𝑙𝑛|2𝑦| − 𝑦𝑙𝑛|0|= y 𝑙𝑛|2𝑦| ∫0 ∫0 (𝑥 + 𝑦)𝑑𝑦𝑑𝑥 = 2 ∫0 (𝑥 1 ∫0 (𝑥 + 𝑦)𝑑𝑦=[𝑥𝑦 + 1 𝑥2 2 2 − ) 𝑑𝑥 =[ 𝑦2 1 2 0 𝑥 1 1 2 2 ] =[x- ]+[0+0]=x- 1 − ] =1/2-1/2=0 2 0