Question
Evaluate the following intregals:
$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$
$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$
✓
Answer
Let $\text{I}=\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$
Consider,
$\text{X}=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)+\text{B}$
$\text{x}=\text{A}(2\text{x}+1)+\text{B}$
$\Rightarrow\text{x}=(2\text{A})\text{x}+\text{A}+\text{B}$
Equating coefficient of like terms
$2\text{A}=1$
$\Rightarrow\text{A}=\frac{1}{2}$
And
$\text{A}+\text{B}=0$
$\Rightarrow\frac{1}{2}+\text{B}=0$
$\Rightarrow\text{B}=-\frac{1}{2}$
$\therefore\text{I}=\int\frac{\big(\frac{1}{2}(2\text{x}+1)-\frac{1}{2}\big)}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$
Consider,
$\text{X}=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)+\text{B}$
$\text{x}=\text{A}(2\text{x}+1)+\text{B}$
$\Rightarrow\text{x}=(2\text{A})\text{x}+\text{A}+\text{B}$
Equating coefficient of like terms
$2\text{A}=1$
$\Rightarrow\text{A}=\frac{1}{2}$
And
$\text{A}+\text{B}=0$
$\Rightarrow\frac{1}{2}+\text{B}=0$
$\Rightarrow\text{B}=-\frac{1}{2}$
$\therefore\text{I}=\int\frac{\big(\frac{1}{2}(2\text{x}+1)-\frac{1}{2}\big)}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$
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