Question

Calcular: $\int\limits {(2x+3)\sqrt{x^{2}+3x+1 } } \, dx$

Answer 1

Resposta:

$\frac{2}{3}*(x^2+3x+1)^\frac{3}{2} + c$

Explicação passo-a-passo:

Resolvendo a primitiva:

Tomando u = $x^2 + 3x + 1$, teremos que:

$du = 2x+3dx\\\\dx = \frac{du}{2x+3}$

Assim temos:

$=\int {(2x+3)*\sqrt{u}} \, \frac{du}{2x+3}\\\\=\int {\sqrt{u}} \, du \\\\=\frac{2}{3} * u^\frac{3}{2} \\\\=\frac{2}{3}*(x^2+3x+1)^\frac{3}{2} + c$

Espero ter ajudado.