Evaluate ∫x3x2+1dx\int x^3 \sqrt{x^2+1} dx∫x3x2+1dx using the substitution u=x2+1u = x^2+1u=x2+1.
13(x2+1)3/2−(x2+1)1/2+C\frac{1}{3}(x^2+1)^{3/2} - (x^2+1)^{1/2} + C31(x2+1)3/2−(x2+1)1/2+C
15(x2+1)5/2−13(x2+1)3/2+C\frac{1}{5}(x^2+1)^{5/2} - \frac{1}{3}(x^2+1)^{3/2} + C51(x2+1)5/2−31(x2+1)3/2+C
13x3(x2+1)3/2+C\frac{1}{3}x^3(x^2+1)^{3/2} + C31x3(x2+1)3/2+C
12(x2+1)5/2+C\frac{1}{2}(x^2+1)^{5/2} + C21(x2+1)5/2+C