Evaluate the following integrals: (x+1) x^2+x+1 dx - Vidyadip

Question

Evaluate the following integrals:
$\int(\text{x}+1)\sqrt{\text{x}^2+\text{x}+1}\text{dx}$

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Answer

Let $\text{I}=\int(\text{x}+1)\sqrt{\text{x}^2+\text{x}+1}\text{dx}$Let $\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)+\mu$
$=\lambda(2\text{x}+1)+\mu$ Equating similar terms, we get, $2\lambda=1\Rightarrow\lambda=\frac{1}{2}$ $\lambda+\mu=1\Rightarrow\mu=\frac{1}{2}$ So, $\text{I}=\int\Big(\frac{1}{2}(2\text{x}+1)+\frac{1}{2}\Big)\sqrt{\text{x}^2+\text{x}+1}\text{dx}$ $=\frac{1}{2}\int(2\text{x}+1)\sqrt{\text{x}^2+\text{x}+1}\text{dx}+\frac{1}{2}\int\sqrt{\text{x}^2+\text{x}+1}\text{dx}$ Let $\text{x}^2+\text{x}+1=\text{t}$ $\Rightarrow(2\text{x}+1)\text{dx = dt}$ $=\frac{1}{2}\int\sqrt{\text{t}}\text{dt}+\frac{1}{2}\int\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt3}{2}\Big)^2}\text{dx}$ $=\frac{1}{2}.\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{1}{2}\begin{Bmatrix}\frac{\big(\text{x}+\frac{1}{2}\big)}{2}\sqrt{\text{x}^2+\text{x}+1}\\+\frac{3}{8}\log\bigg|\Big(\text{x}+\frac{1}{2}\Big)+\sqrt{\text{x}^2+\text{x}+1}\bigg|\end{Bmatrix}+\text{C}$ $\Rightarrow\text{I}=\frac{1}{3}(\text{x}^2+\text{x}+1)^{\frac{3}{2}}+\frac{1}{8}(2\text{x}+1)\sqrt{\text{x}^2+\text{x}+1}\\+\frac{3}{16}\log\bigg|\Big(\text{x}+\frac{1}{2}\Big)+\sqrt{\text{x}^2+\text{x}+1}\bigg|+\text{C}$ Hence, $\text{I}=\frac{1}{3}(\text{x}^2+\text{x}+1)^{\frac{3}{2}}+\frac{1}{8}(2\text{x}+1)\sqrt{\text{x}^2+\text{x}+1}\\+\frac{3}{16}\log\bigg|\Big(\text{x}+\frac{1}{2}\Big)+\sqrt{\text{x}^2+\text{x}+1}\bigg|+\text{C}$

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