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

Question

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

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Answer

Let $\text{I}=\int\text{x}\sqrt{\text{x}^2+\text{x}}\text{dx}$
Let $\text{x}=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x})+\mu$
$=\lambda(2\text{x}+1)+\mu$
Equating similar terms, we get,
$2\lambda=1\Rightarrow\lambda=\frac{1}{2}$
$\lambda+\mu=0\Rightarrow\mu=-\frac{1}{2}$
So,
$\text{I}=\frac{1}{2}\int\Big(\frac{1}{2}(2\text{x}+1)-\frac{1}{2}\Big)\sqrt{\text{x}^2+\text{x}}\text{dx}$
$=\frac{1}{2}\int(2\text{x}+1)\sqrt{\text{x}^2+\text{x}}-\frac{1}{2}\int\sqrt{\text{x}^2+\text{x}}\text{dx}$
Let $\text{x}^2+\text{x}=\text{t}$
$\Rightarrow(2\text{x}+1)\text{dx = dt}$
So,
$\text{I}=\frac{1}{2}\int\sqrt{\text{t}}\text{dt}-\frac{1}{2}\int\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2}$
$\text{I}=\frac{1}{2}.\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{2}\begin{Bmatrix}\frac{\big(\text{x}+\frac{1}{2}\big)}{2}\sqrt{\text{x}^2+\text{x}}\\-\frac{1}{8}\log\Big|\Big(\text{x}+\frac{1}{2}\Big)+\sqrt{\text{x}^2+\text{x}}\Big|\end{Bmatrix}+\text{C}$
Hence,
$\text{I}=\frac{1}{3}(\text{x}^2+\text{x})^{\frac{3}{2}}-\frac{1}{8}{\Big(\text{x}+\frac{1}{2}\Big)}{2}\sqrt{\text{x}^2+\text{x}}\\+\frac{1}{16}\log\Big|\Big(\text{x}+\frac{1}{2}\Big)+\sqrt{\text{x}^2+\text{x}}\Big|+\text{C}$

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