The integral ∫1/(3x+4)³ from a(1) to b(∞) is ∫₁⁰⁰[1/(3x+4)³]dx = 1/6
To find the integral ∫1/(3x+4)³ from a(1) to b(∞), we proceed as follows
Since we have the integral ∫1/(3x+4)³ from a(1) to b(∞). It is re-written as
∫₁⁰⁰[1/(3x+4)³]dx
Now, let u = 3x + 4
Differentiating, we have that
du/dx = d(3x + 4)/dx
= d3x/dx + d4/dx
= 3dx/dx + 0
= 3
du/dx = 3
dx = du/3
So, substituting u and dx into the integral equation, we have that
∫₁⁰⁰[1/(3x+4)³]dx = ∫₁⁰⁰[1/u³]du/3
= 1/3∫₁⁰⁰[1/u³]du
= 1/3[u⁻³⁺¹/(-3 + 1)]₁⁰⁰
= 1/3[u⁻²/(-2)]₁⁰⁰
= -1/6[u⁻²]₁⁰⁰
= -1/6[1/u²]₁⁰⁰
= -1/6[1/∞² - 1/1²]
= -1/6[1/∞ - 1/1]
= -1/6[0 - 1]
= -1/6[-1]
= 1/6
So, ∫₁⁰⁰[1/(3x+4)³]dx = 1/6