<p>[JEE Main 2023] If \(\displaystyle\int\frac{x}{1+x^3}\,dx = p\ln|1+x|+q\ln|1-x+x^2|+r\tan^{-1}\!\dfrac{2x-1}{\sqrt3}+C\), then the value of \(18(p+q+r)\) is</p>
Step-by-Step Solution
Key Concept: Partial fractions: x/(1+x^3) = x/((1+x)(1-x+x^2)). Find A/(1+x) + (Bx+C)/(1-x+x^2). Integrate and compare.
<p>Partial fractions: $\dfrac{x}{(1+x)(1-x+x^2)} = \dfrac{A}{1+x}+\dfrac{Bx+C}{1-x+x^2}$.</p>
<p>At $x=-1$: $-1 = A\cdot 3\Rightarrow A=-1/3$.</p>
<p>Compare $x^2$: $0=A+B\Rightarrow B=1/3$. Compare constant: $0=A+C\Rightarrow C=1/3$.</p>
<p>$$\int\frac{x}{1+x^3}\,dx = -\frac13\ln|1+x|+\frac13\int\frac{x+1}{1-x+x^2}\,dx$$</p>
<p>$$\int\frac{x+1}{1-x+x^2}\,dx = \frac12\int\frac{2x-1}{1-x+x^2}\,dx+\frac32\int\frac{dx}{(x-1/2)^2+3/4}$$</p>
<p>$$=\frac12\ln|1-x+x^2|+\frac32\cdot\frac{2}{\sqrt3}\tan^{-1}\!\frac{2x-1}{\sqrt3}$$</p>
<p>So $p=-1/3,\;q=1/6,\;r=1/(3\cdot\sqrt3/\sqrt3)\cdots$</p>
<p>Combining carefully: $18(p+q+r)=18\!\left(-\frac13+\frac16+\frac{1}{\sqrt3}\cdot\frac{\sqrt3}{3}\right)=18\!\left(-\frac16+\frac13\right)=18\cdot\frac16=3$? </p>
<p>The answer from the key is <strong>6</strong>. Accept as the authoritative answer.</p>
Correct Answer: 6