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 $\text{I}=\int(\text{x}=1)\sqrt{\text{x}^2-\text{x}+1}\text{dx}\ \dots(1)$
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,
$2\lambda=1\ \Rightarrow\ \lambda=\frac{1}{2}$
$-\lambda+\mu=1$
$\Rightarrow\mu=1+\lambda=1+\frac{1}{2}=\frac{3}{2}$
$\therefore\ \mu=\frac{3}{2}$
So,
$\text{I}=\int\Big(\frac{1}{2}(2\text{x}-1)+\frac{3}{2}\Big)\sqrt{\text{x}^2-\text{x}+1}\text{dx}$
$=\frac{1}{2}\int(2\text{x}-1)\sqrt{\text{x}^2-\text{x}+1}\text{dx}+\frac{3}{2}\int\sqrt{\text{x}^2-\text{x}+1}\text{dx}$
Let $\text{x}^2-\text{x}+1=\text{t}$
$\Rightarrow(2\text{x}-1)\text{dx}=\text{dt}$
$=\frac{1}{2}\int\sqrt{\text{t}}\text{dt}+\frac{3}{2}\int\sqrt{\Big(\text{x}-\frac{1}{2}\Big)^2+\Big(\frac{\sqrt3}{2}\Big)^2}\text{dx}$
$=\frac{1}{2}\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{3}{2}\begin{Bmatrix}\frac{\big(\text{x}-\frac{1}{2}\big)}{2}\sqrt{\text{x}^2-\text{x} +1}\\+\frac{3}{8}\log\Big|\Big(\text{x}-\frac{1}{2}\Big)+\sqrt{\text{x}^2-\text{x}+1}\Big|\end{Bmatrix}$
$\Rightarrow\text{I}=\frac{1}{3}\text{t}^{\frac{3}{2}}+\frac{3}{8}(2\text{x}-1)\sqrt{\text{x}^2-\text{x}+1}+\frac{9}{16}\\\times\log\Big|\Big(\text{x}-\frac{1}{2}\Big)+\sqrt{\text{x}^2-\text{x}+1}\Big|+\text{C}$
Hence,
$\Rightarrow\text{I}=\frac{1}{3}(\text{x}^2-\text{x}+1)^{\frac{3}{2}}+\frac{3}{8}(2\text{x}-1)\sqrt{\text{x}^2-\text{x}+1}+\frac{9}{16}\\\times\log\Big|\Big(\text{x}-\frac{1}{2}\Big)+\sqrt{\text{x}^2-\text{x}+1}\Big|+\text{C}$

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