Question

Integral por Substituição :

∫√x²+1 x dx ???

Answer 1

$\displaystyle\int\sqrt{x^{2}+1}xdx$

Fazendo $u=x^{2}+1$ e derivando u, encontramos

$du=2xdx~~\longrightarrow~~xdx=\frac{1}{2}du$

Então:

$\displaystyle\int\sqrt{x^{2}+1}xdx=\int\sqrt{u}\dfrac{1}{2}du\\\\\\\int\sqrt{x^{2}+1}xdx=\dfrac{1}{2}\int u^{\frac{1}{2}}du\\\\\\\int\sqrt{x^{2}+1}xdx=\dfrac{1}{2}\dfrac{u^{\frac{1}{2}+1}}{(\frac{1}{2}+1)}+constante\\\\\\\int\sqrt{x^{2}+1}xdx=\dfrac{1}{2}\cdot\dfrac{u^{\frac{3}{2}}}{(\frac{3}{2})}+constante\\\\\\\int\sqrt{x^{2}+1}xdx=\dfrac{1}{2\cdot(\frac{3}{2})}(x^{2}+1)^{\frac{3}{2}}+constante\\\\\\\boxed{\boxed{\int\sqrt{x^{2}+1}xdx=\dfrac{1}{3}(x^{2}+1)^{\frac{3}{2}}+constante}}$