Chapter 5 - The Integral - 5.5 The Fundamental Theorem of Calculus, Part II - Exercises - Page 263: 29

$\frac{2x^3}{x^2+1}.$

By making use of Theorem 1 (FTC II), we have $$ \frac{d}{d x} \int_{0}^{x^{2}} \frac{t d t}{t+1}=\left(\frac{d}{d u} \int_{0}^{u} \frac{t d t}{t+1}\right)\frac{du}{dx}=\frac{x^2}{x^2+1} (2x)\\ =\frac{2x^3}{x^2+1}. $$