Evaluate the integral: I = integral (x³)/(sqrt(x²+2)) dx

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

Step 1 · Substitute

Let $w=x^2+2$, so $dw=2x\,dx$ and $x^2=w-2$:

$$ I = \int\frac{x^2\cdot x\,dx}{\sqrt{x^2+2}} = \frac12\int\frac{(w-2)}{\sqrt w}\,dw = \frac12\int\left(w^{1/2}-2w^{-1/2}\right)dw. $$

Step 2 · Integrate

$$ = \frac12\left[\frac{2w^{3/2}}3 - 4w^{1/2}\right] + C = \frac{w^{3/2}}3 - 2w^{1/2} + C. $$

Step 3 · Back-substitute

With $w=x^2+2$:

$$ I = \frac{(x^2+2)^{3/2}}3 - 2\sqrt{x^2+2} + C = \frac{(x^2+2)\sqrt{x^2+2}}3 - 2\sqrt{x^2+2} + C. $$

Combining over a common denominator of $3$:

Summary Table

Additional Notes

• The domain is all real $x$, since $x^2+2>0$ everywhere.
• The result was verified by differentiation: $\tfrac{d}{dx}\left[\tfrac{(x^2-4)\sqrt{x^2+2}}3\right] = \tfrac{2x\sqrt{x^2+2}}3+\tfrac{x(x^2-4)}{3\sqrt{x^2+2}} = \tfrac{2x(x^2+2)+x(x^2-4)}{3\sqrt{x^2+2}} = \tfrac{3x^3}{3\sqrt{x^2+2}}=\tfrac{x^3}{\sqrt{x^2+2}}$, matching the integrand exactly.