Question

Integrate the function: (4x + 2)$\sqrt {x^2+x+1}$

Answer 1

Let x2+x+1 = t

∴ (2x + 1)dx = dt

$\int$(4x + 2)$\sqrt {x^2+x+1}$ dx

= $\int$2$\sqrt t$ dt

= 2 $\int \sqrt t$ dt

=$2\bigg(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\bigg)+C$

=$\frac{4}{3}(x^2+x+1)^{\frac{3}{2}}+C$