Determine whether each integral is convergent or divergent.

1 - 1 / (x-2) ^3/2 dx, limits (infinite to 3)

2 - (1/3-4x) dx, limits (0 to - infinite)

3 - e^ (-5p) dx, limits (infinite to 2)

4 - (x^2 / (sqrt (1+x^3))) dx, limits (infinite to 0)

5 - lnx/x dx, limits (infinite to 1)

6 - 1 / (x^2 + x) dx, limits (infinite to 1)

7 - 3/x^5 dx, limits (1 to 0)

8 - dx / (x+2) ^1/4, limits (14 to - 2)

9 - 1 / (x-1) ^1/3, limits (9 to 0)

10 - e^x / ((e^x) - 1), limits (1 to - 1)

11 - z^2 lnz dz, limits (2 to 0)

12 - s = (x, y)

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

Divergers