Answer:
Step-by-step explanation: 1) ∫ 1/( x - 2)∧3/2 dx, Limits ⇒ infinite to 3
∫( x - 2)∧-3/2 = -2/( x - 2)∧1/2 + c
Limit⇒ infinite to 3: -2/(3 - 2) = -2 ( convergent)
2) ∫1/( 3 - 4x) dx, Limits⇒ 0 to infinite
Let u = 3 - 4x
du = -4 dx
∴ dx = -1/4du
Hence, -1/4∫1/udu = - 1/4ln u + c = -1/4ln(3 - 4x) + c
Limits⇒ 0 to infinite: -1/4㏑(3 - 0) = -1/4 ㏒3 (convergent)
3) ∫e∧(-5x)dx, Limits⇒ infinite to 2
Let u = -5x
du = -5dx
∴ dx = -1/5du
-1/5∫e∧-u = -1/5e∧-5x + c
Limits⇒ infinite to 2: -1/5e∧-5(2) = -1/5e∧-10 (divergent)
4) ∫x²/√(1 + x³)dx, Limits⇒ infinite to 0
Let u = 1 + x³
du = 3x²dx
∴ dx = 1/3x²du
1/3x²∫x²/u∧1/2du = 1/3㏑u∧1/2 + c = 1/3㏑(√1 + x²) + c
Limits⇒infinite to 0: 1/3㏒0 = infinite (divergent)
5)
6)∫1/(x² + x)dx, Limits⇒ infinite to 1
㏑(x² + x) + c
Limits⇒ infinite to 1: ㏒(1 + 1) = ㏒ 2 (divergent)
7) ∫3/x∧5dx = ∫3x∧-5dx, Limits⇒ (1 to 0)
- 3/4x∧-4 + c
Limits ( 1 to 0): -3/4(1)∧-4 - (-3/4(0)) = -3/4 + 0 = -3/4 (convergent)
8)∫1/(x + 2)∧1/4 dx, Limits (14 to -2)
Let u = x + 2
du = dx
∫1/u∧1/4 du = ∫u∧-1/4
3/4u∧3/4 = 3(x + 2)∧3/4/4 + c
Limits (14 to -2): 3(14 + 2)∧3/4/4 - (3(-2 + 2)∧3/4/4) = 3(16)∧3/4/4 - 0
3(2)³/4 = 3 X 8/4 = 3 X 2 = 6 (convergent)
9) ∫1/(x - 1)∧1/3dx, Limits (9 to 0)
Let u = x - 1
du = dx
∫1/u∧1/3 du = ∫u∧-1/3du = 2u∧2/3/3 + c = 2(x - 1)∧2/3/3 + c
Limits (9 to 0): 2(9 -1)∧2/3/3 - 2(0 - 1)∧2/3/3 = 2(8)∧2/3 - 2(-1)∧2/3/3
2(2)²/3 - 2/3 = 8/3 - 2/3 = 6/3 = 2(convergent)
10) ∫e∧x/(e∧x - 1), Limits (1 to -1)
Let u = e∧x - 1
du = e∧x dx
∴ dx = 1/e∧x du
1/e∧x∫e∧x/u du
∫1/u = ㏑u du + c = ㏒(e∧x - 1) + c
Limits (1 to -1): ㏒(e - 1) - ㏒e∧-1 - 1
㏒(e - e∧-1) (divergent)
11) ∫z∧2㏑z dz, Limits (2 to 0) (divergent)
12) Diververgent