Integrate the function (4x+2)sqrt{x^{2}+x+1}. | vedclass

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(A) Let $I = \int (4x+2)\sqrt{x^{2}+x+1} \, dx$. Substitute $t = x^{2}+x+1$. Then,$dt = (2x+1) \, dx$,which implies $2 \, dt = (4x+2) \, dx$. Substitu

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int (4x+2)\sqrt{x^{2}+x+1} \, dx$. Substitute $t = x^{2}+x+1$. Then,$dt = (2x+1) \, dx$,which implies $2 \, dt = (4x+2) \, dx$. Substituting these into the integral: $I = \int 2\sqrt{t} \, dt = 2 \int t^{\frac{1}{2}} \, dt$. Using the power rule $\int t^{n} \, dt = \frac{t^{n+1}}{n+1} + C$: $I = 2 \left( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} ight) + C = 2 	imes \frac{2}{3} t^{\frac{3}{2}} + C = \frac{4}{3} t^{\frac{3}{2}} + C$. Substituting back $t = x^{2}+x+1$: $I = \frac{4}{3}(x^{2}+x+1)^{\frac{3}{2}} + C$,where $C$ is an arbitrary constant.

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

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