$\int \frac{\left(x+\sqrt{1+x^2}\right)^{15}}{\sqrt{1+x^2}} d x=$

Question

Professor PC Thomas Classes · Answer

Substitute $x+\sqrt{1+x^2} = t \implies dt = (1+\frac{x}{\sqrt{1+x^2}})dx = \frac{t}{\sqrt{1+x^2}}dx \implies \int t^{14} dt = \frac{t^{15}}{15} + c$.