MAT1613 Assignment 2 2021

MAT1613 ASSIGNMENT 2 2021



Question 1



𝑒 √𝑥
𝐼=∫ 𝑑𝑥
√𝑥


1
𝐼 = ∫ 𝑒 √𝑥 𝑑𝑥
√𝑥


𝐿𝑒𝑡 𝑢 = √𝑥

𝑢 = 𝑥 1⁄2
1
𝑑𝑢 = 𝑥 1⁄2−1 𝑑𝑥
2
1
𝑑𝑢 = 𝑥 −1⁄2 𝑑𝑥
2
1
𝑑𝑢 = 𝑑𝑥
2𝑥 1⁄2
1
𝑑𝑢 = 𝑑𝑥
2√𝑥
1
2𝑑𝑢 = 𝑑𝑥
√𝑥


1
𝐼 = ∫ 𝑒 √𝑥 𝑑𝑥
√𝑥

𝐼 = ∫ 𝑒 𝑢 2𝑑𝑢


𝐼 = 2 ∫ 𝑒 𝑢 𝑑𝑢

𝐼 = 2𝑒 𝑢 + 𝑐

𝐼 = 2𝑒 √𝑥 + 𝑐



Question 1 TWO

,Question 2


8𝑑𝑥
𝐼=∫
𝑥(𝑙𝑛𝑥 + 1)3⁄2
8 1
𝐼=∫ 𝑑𝑥
(𝑙𝑛𝑥 + 1)3⁄2 𝑥



𝐿𝑒𝑡 𝑢 = 𝑙𝑛𝑥 + 1
1
𝑑𝑢 = 𝑑𝑥
𝑥


8 1
𝐼=∫ 3 ⁄2
𝑑𝑥
(𝑙𝑛𝑥 + 1) 𝑥
8
𝐼=∫ 𝑑𝑢
𝑢 3 ⁄2

𝐼 = ∫ 8𝑢−3⁄2 𝑑𝑢


𝐼 = 8 ∫ 𝑢−3⁄2 𝑑𝑢



1
𝐼 = 8[ 𝑢−3⁄2+1 ] + 𝑐
−3⁄2 + 1


1
𝐼 = 8[ 𝑢−3⁄2+2⁄2 ] + 𝑐
−3⁄2 + 2⁄2


1
𝐼 = 8[ 𝑢(−3+2)⁄2 ] + 𝑐
(−3 + 2)⁄2


1
𝐼 = 8[ 𝑢−1⁄2 ] + 𝑐
−1⁄2



𝐼 = 8[−2𝑢−1⁄2 ] + 𝑐



𝐼 = −16[𝑢−1⁄2 ] + 𝑐

,𝐼 = −16(𝑙𝑛𝑥 + 1)−1⁄2 + 𝑐


16
𝐼=− +𝑐
(𝑙𝑛𝑥 + 1)1⁄2


16
𝐼=− +𝑐
√𝑙𝑛𝑥 + 1


−16
𝐼= +𝑐
√𝑙𝑛𝑥 + 1


Question 2 FOUR



Question 3


3𝑡 + 2
𝐼=∫ 𝑑𝑡
𝑡+1


𝐿𝑒𝑡 𝑢 = 𝑡 + 1
𝑑𝑢 = 𝑑𝑡



𝑢 =𝑡+1
⇒𝑡 =𝑢−1


3𝑡 + 2
𝐼=∫ 𝑑𝑡
𝑡+1


3(𝑢 − 1) + 2
𝐼=∫ 𝑑𝑢
𝑢


3𝑢 − 3 + 2
𝐼=∫ 𝑑𝑢
𝑢


3𝑢 − 1
𝐼=∫ 𝑑𝑢
𝑢

, 3𝑢 1
𝐼 = ∫[ − ] 𝑑𝑢
𝑢 𝑢


1
𝐼 = ∫ [3 − ] 𝑑𝑢
𝑢


1
𝐼 = ∫ 3 𝑑𝑢 − ∫ 𝑑𝑢
𝑢


𝐼 = 3𝑢 − 𝑙𝑛|𝑢| + 𝑐



𝐼 = 3(𝑡 + 1) − 𝑙𝑛|𝑡 + 1| + 𝑐



𝐼 = 3𝑡 + 3 − 𝑙𝑛|𝑡 + 1| + 𝑐



𝐼 = 3𝑡 − 𝑙𝑛|𝑡 + 1| + 𝑐 + 3



𝐼 = 3𝑡 − 𝑙𝑛|𝑡 + 1| + (𝑐 + 3)



𝐼 = 3𝑡 − 𝑙𝑛|𝑡 + 1| + 𝑐2



Question 3 ONE



Question 4



5𝑡√𝑡
𝐼=∫ 𝑑𝑡
1 + 𝑡5



5𝑡𝑡 1⁄2
𝐼=∫ 𝑑𝑡
1 + 𝑡5



5𝑡1 𝑡 1⁄2
𝐼=∫ 𝑑𝑡
1 + 𝑡5