Question

The value of \(\displaystyle\int_{0}^{2}\frac{x^2}{(x^3+1)^2}dx\) is equal to

• A. \(\frac{1}{27}\)
• B. \(\frac{5}{27}\)
• C. \(\frac{7}{27}\)
• D. \(\frac{8}{27}\)
• E. \(\frac{1}{3}\)

Answer 1

The Correct Option is D

We are asked to find the value of the integral \( \int_0^2 \frac{x^2}{(x^3 + 1)^2} \, dx \). 

We will solve this using substitution. Let:

\( u = x^3 + 1 \),

so that \( du = 3x^2 \, dx \), or equivalently, \( \frac{du}{3} = x^2 \, dx \).

Now, the limits of integration change accordingly. When \( x = 0 \), \( u = 0^3 + 1 = 1 \), and when \( x = 2 \), \( u = 2^3 + 1 = 9 \).

The integral becomes:

\( \int_1^9 \frac{1}{3u^2} \, du \).

Now, we can integrate \( \frac{1}{u^2} \), which gives \( -\frac{1}{u} \). So the integral becomes:

\( \frac{1}{3} \left[ -\frac{1}{u} \right]_1^9 = \frac{1}{3} \left( -\frac{1}{9} + 1 \right) \).

This simplifies to:

\( \frac{1}{3} \left( \frac{8}{9} \right) = \frac{8}{27} \).

The correct answer is \( \frac{8}{27} \).