Question

The value of \(\displaystyle \int_{0}^{1} \frac{t}{(t+1)^3}\,dt\) is equal to:

• A. \(\frac{1}{8}\)
• B. \(\frac{3}{8}\)
• C. \(\frac{5}{8}\)
• D. \(\frac{7}{8}\)
• E. \(\frac{1}{4}\)

Hint:

Manipulating the numerator to look like the denominator (\(t+1-1\)) is almost always faster than using substitution in rational integrals.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We can solve this definite integral by adjusting the numerator to match the base of the denominator.

Step 2: Key Formula or Approach:
Write \(t = (t + 1) - 1\).

Step 3: Detailed Explanation:
\[ \int_0^1 \frac{t + 1 - 1}{(t + 1)^3} dt = \int_0^1 \left[ \frac{1}{(t + 1)^2} - \frac{1}{(t + 1)^3} \right] dt \]
Integrate each term:
\[ = \left[ -\frac{1}{t + 1} + \frac{1}{2(t + 1)^2} \right]_0^1 \]
Apply upper limit (t = 1):
\[ = -\frac{1}{2} + \frac{1}{2(2)^2} = -\frac{1}{2} + \frac{1}{8} = -\frac{3}{8} \]
Apply lower limit (t = 0):
\[ = -\frac{1}{1} + \frac{1}{2(1)^2} = -1 + \frac{1}{2} = -\frac{1}{2} \]
Calculate the difference:
\[ \text{Value} = -\frac{3}{8} - (-\frac{1}{2}) = -\frac{3}{8} + \frac{4}{8} = \frac{1}{8} \]

Step 4: Final Answer:
The value of the integral is 1/8.