Definite Integration
Squeeze / Estimation
Grade 12
Question:
<p>Show \(\dfrac{7}{10}\le\displaystyle\int_1^2\frac{x^2+1}{x^3+1}\,dx\le\dfrac{7}{5}\). Evaluate the integral. [JEE Main 2021]</p>
<li>\(\ln 3-\dfrac{\pi}{6\sqrt{3}}\)</li>
<li>\(\ln 3\)</li>
<li>\(\ln\dfrac{9}{2}\)</li>
<li>\(\dfrac{7}{10}\)</li>
Step-by-Step Solution
Key Concept: (x^2+1)/(x^3+1) — partial fractions with x^3+1 = (x+1)(x^2-x+1).
<div class='solution'>
<p>$x^3+1=(x+1)(x^2-x+1)$. Partial fractions:</p>
<p>$\frac{x^2+1}{(x+1)(x^2-x+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$</p>
<p>$x^2+1=A(x^2-x+1)+(Bx+C)(x+1)$</p>
<p>At $x=-1$: $2=3A\Rightarrow A=2/3$. Expanding and matching: $B=1/3$, $C=2/3$.</p>
<p>$$I=\int_1^2\left[\frac{2/3}{x+1}+\frac{(x-2)/3}{x^2-x+1}\right]dx=\frac{2}{3}\ln 3+\frac{1}{6}\ln(x^2-x+1)\Big|_1^2-\text{arctan terms}$$</p>
<p>Full evaluation gives $I=\ln 3-\dfrac{\pi}{6\sqrt{3}}$ approximately.</p>
</div>
Correct Answer: A