Question

Evaluate the integral: \[ \int_0^1 \frac{t}{(t + 1)^3} \, dt \]

• A. \( \frac{1}{8} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{1}{16} \)

Hint:

When performing substitution, always remember to change both the differential and the integrand in terms of the new variable. This simplifies the integral.

Answer 1

The Correct Option is A

We are tasked with evaluating the integral: \[ I = \int_0^1 \frac{t}{(t + 1)^3} \, dt \] Step 1: Use substitution.
Let’s make the substitution: \[ u = t + 1 \] Therefore, \[ du = dt \quad \text{and} \quad t = u - 1 \] The limits of integration change as follows:
- When \( t = 0 \), \( u = 1 \),
- When \( t = 1 \), \( u = 2 \).
Now, substitute into the integral: \[ I = \int_1^2 \frac{u - 1}{u^3} \, du \]
Step 2: Simplify the integrand.
We can split the fraction: \[ I = \int_1^2 \left( \frac{u}{u^3} - \frac{1}{u^3} \right) du \] \[ I = \int_1^2 \left( \frac{1}{u^2} - \frac{1}{u^3} \right) du \]
Step 3: Integrate term by term.
Now, integrate each term: \[ \int \frac{1}{u^2} \, du = -\frac{1}{u} \] \[ \int \frac{1}{u^3} \, du = -\frac{1}{2u^2} \] Thus, the integral becomes: \[ I = \left[ -\frac{1}{u} + \frac{1}{2u^2} \right]_1^2 \]
Step 4: Evaluate the definite integral.
Now, substitute the limits of integration: \[ I = \left( -\frac{1}{2} + \frac{1}{2(2)^2} \right) - \left( -\frac{1}{1} + \frac{1}{2(1)^2} \right) \] \[ I = \left( -\frac{1}{2} + \frac{1}{8} \right) - \left( -1 + \frac{1}{2} \right) \] \[ I = \left( -\frac{1}{2} + \frac{1}{8} \right) - \left( -\frac{1}{2} \right) \] \[ I = \frac{1}{8} \] Thus, the value of the integral is: \[ \boxed{\frac{1}{8}} \]