Question

Calcule $\int (\sqrt{1 + x^2} + 2x) dx$

Answer 1

$\displaystyle I=\int \sqrt{1+x^2}\,dx +\int 2x\,dx\\ \\ I=\int \sqrt{1+x^2}\,dx +x^2$

calculemos

$\displaystyle J=\int \sqrt{1+x^2}\,dx\\ \\ \text{Seja }x=\sinh u\to dx = \cosh u \,du\\ \\ J=\int \sqrt{1+\sinh^2u}\,\cosh u \,du\\ \\ J=\int \cosh u\cdot \cosh u \,du\\ \\ J=\int \cosh^2 u \,du\\ \\ J=\int \dfrac{\cosh(2u) + 2}{4} \,du$

$\displaystyle J=\dfrac{1}{4}\int \cosh(2u) \,du + \int \dfrac{1}{2} \,du\\ \\ J=\dfrac{1}{8}\sinh(2u) +\dfrac{u}{2}\\ \\ J=\dfrac{1}{2}\sinh u\cosh u +\dfrac{u}{2}\\ \\ J=\dfrac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\sinh^{-1}x \\ \\ J=\dfrac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\ln \left|x+\sqrt{1+x^2}\right|$

finalmente

$\boxed{I=\dfrac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\ln \left|x+\sqrt{1+x^2}\right|+x^2+C}$