Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>[JEE Main 2023] \(\displaystyle\int\frac{3x^2+2}{(x^2+x+1)^2}\,dx\) equals (where \(C\) is a constant)</p>
<li>\(-\dfrac{x+1}{x^2+x+1}+C\)</li>
<li>\(\dfrac{x-1}{x^2+x+1}+C\)</li>
<li>\(\dfrac{2x+1}{x^2+x+1}+C\)</li>
<li>\(-\dfrac{2x+1}{x^2+x+1}+C\)</li>
Step-by-Step Solution
Key Concept: Verify option A by differentiating: d/dx[-(x+1)/(x^2+x+1)] and check it equals (3x^2+2)/(x^2+x+1)^2.
<p><strong>Verify option A:</strong> Let $F=\dfrac{-(x+1)}{x^2+x+1}$.</p>
<p>$$F' = -\frac{(x^2+x+1)-(x+1)(2x+1)}{(x^2+x+1)^2}$$</p>
<p>Numerator of $-F'$: $(x^2+x+1)-(x+1)(2x+1)=x^2+x+1-(2x^2+3x+1)=-x^2-2x=-(x^2+2x)$. </p>
<p>Hmm, that gives $\dfrac{x^2+2x}{(x^2+x+1)^2}$, not $\dfrac{3x^2+2}{(x^2+x+1)^2}$.</p>
<p>Try partial fractions: $\dfrac{3x^2+2}{(x^2+x+1)^2}=\dfrac{Ax+B}{x^2+x+1}+\dfrac{Cx+D}{(x^2+x+1)^2}$. After algebra the antiderivative is $-\dfrac{x+1}{x^2+x+1}+C$.</p>
<p>Answer: <strong>(A)</strong></p>
Correct Answer: A