If $$\int\ x^2\tan^{-1}xdx=Ax^3\tan^{-1}x+Bx^2+C\ln\left(1+x^2\right)+D$$, then $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=$$

Evaluate the integral using integration by parts by setting $$u = \tan^{-1} x$$ and $$dv = x^2 \, dx$$:

$$\int x^2 \tan^{-1} x \, dx = \frac{x^3}{3} \tan^{-1} x - \int \frac{x^3}{3(1 + x^2)} \, dx$$

Simplify the rational function using polynomial adjustment: $$\frac{x^3}{1 + x^2} = x - \frac{x}{1 + x^2}$$

Integrate the separated terms: $$\int \left( x - \frac{x}{1 + x^2} \right) dx = \frac{x^2}{2} - \frac{1}{2} \ln(1 + x^2)$$

Substitute this result back into the main expression:

$$\int x^2 \tan^{-1} x \, dx = \frac{1}{3} x^3 \tan^{-1} x - \frac{1}{6} x^2 + \frac{1}{6} \ln(1 + x^2) + D$$

Compare coefficients with the given standard form $$A x^3 \tan^{-1} x + B x^2 + C \ln(1 + x^2) + D$$:

• $$A = \frac{1}{3}$$
  
• $$B = -\frac{1}{6}$$
  
• $$C = \frac{1}{6}$$
  

Compute the reciprocal values to find the final sum:

• $$\frac{1}{A} = 3$$
  
• $$\frac{1}{B} = -6$$
  
• $$\frac{1}{C} = 6$$
  

Sum the reciprocals: $$\frac{1}{A} + \frac{1}{B} + \frac{1}{C} = 3 - 6 + 6 = 3$$