Evaluate the integral: I = integral (x^4)/(x²+3) dx

Evaluate the integral: $$ I = \int \frac{x^4}{x^2+3}\,dx. $$

Step 1 · Polynomial long division

$$ \frac{x^4}{x^2+3} = x^2-3+\frac{9}{x^2+3}. $$

Step 2 · Integrate the polynomial part

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

Step 3 · Integrate the remainder

Using the standard formula $\int dx/(x^2+a^2)=(1/a)\arctan(x/a)$ with $a=\sqrt3$:

$$ 9\int \frac{dx}{x^2+3} = \frac{9}{\sqrt3}\arctan\left(\frac{x}{\sqrt3}\right) = 3\sqrt3\arctan\left(\frac{x}{\sqrt3}\right). $$

Step 4 · Combine

$$ I = \frac{x^3}{3}-3x + 3\sqrt3\arctan\left(\frac{x}{\sqrt3}\right) + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$, since $x^2+3>0$ everywhere.
• This is a cleaner case than Problems 231 and 233–235 since the denominator $x^2+3$ is already irreducible with no linear term, so no completing-the-square step is needed before applying the arctangent formula directly.
• The result can be checked by differentiating $\tfrac{x^3}3-3x+3\sqrt3\arctan(x/\sqrt3)$: this gives $x^2-3+\tfrac{3\sqrt3\cdot(1/\sqrt3)}{1+x^2/3}=x^2-3+\tfrac{3}{(3+x^2)/3}=x^2-3+\tfrac{9}{x^2+3}=\tfrac{(x^2-3)(x^2+3)+9}{x^2+3}=\tfrac{x^4-9+9}{x^2+3}=\tfrac{x^4}{x^2+3}$, matching the integrand.