Question
Evaluate the following definite integrals:
$\int_{1}^\limits{4}\frac{\text{x}^2+\text{x}}{\sqrt{2\text{x}+1}}\text{ dx}$
$\int_{1}^\limits{4}\frac{\text{x}^2+\text{x}}{\sqrt{2\text{x}+1}}\text{ dx}$
✓
Answer
Let $\text{I}=\int_{1}^\limits{4}\frac{\text{x}^2+\text{x}}{\sqrt{2\text{x}+1}}\text{ dx}$
Let $2\text{x}+1=\text{u}$
$\Rightarrow\text{x}=\frac{\text{u}-1}{2}$
$\Rightarrow\text{dx}=\frac{\text{du}}{2}$
$\therefore\ \text{I}=\int\frac{\big(\frac{\text{u}-1}{2}\big)^2+\frac{\text{u}-1}{2}}{\sqrt{\text{u}}}\frac{\text{du}}{2}$
$\Rightarrow\text{I}=\frac{1}{8}\int\frac{\text{u}^2+1-2\text{u}+2\text{u}-2}{\sqrt{\text{u}}}\text{ du}$
$\Rightarrow\text{I}=\frac{1}{8}\int\frac{(\text{u}^2-1)}{\sqrt{\text{u}}}\text{ du}$
$\Rightarrow\text{I}=\frac{1}{8}\int\Big(\text{u}^{\frac{3}{2}}-\text{u}^{-\frac{1}{2}}\Big)\text{du}$
$\Rightarrow\text{I}=\frac{1}{8}\bigg[\frac{2\text{u}^{\frac{5}{2}}}{5}-\frac{2\text{u}^{\frac{1}{2}}}{1}\bigg]$
$\Rightarrow\text{I}=\frac{1}{8}\Big[\frac{2}{5}\times243-6-\frac{2}{5}\times9\sqrt{3}+2\sqrt{3}\Big]$
$\Rightarrow\text{I}=\frac{1}{8}\Big[\frac{456}{5}-\frac{8\sqrt{3}}{5}\Big]$
$\Rightarrow\text{I}=\frac{57-\sqrt{3}}{5}$
Let $2\text{x}+1=\text{u}$
$\Rightarrow\text{x}=\frac{\text{u}-1}{2}$
$\Rightarrow\text{dx}=\frac{\text{du}}{2}$
$\therefore\ \text{I}=\int\frac{\big(\frac{\text{u}-1}{2}\big)^2+\frac{\text{u}-1}{2}}{\sqrt{\text{u}}}\frac{\text{du}}{2}$
$\Rightarrow\text{I}=\frac{1}{8}\int\frac{\text{u}^2+1-2\text{u}+2\text{u}-2}{\sqrt{\text{u}}}\text{ du}$
$\Rightarrow\text{I}=\frac{1}{8}\int\frac{(\text{u}^2-1)}{\sqrt{\text{u}}}\text{ du}$
$\Rightarrow\text{I}=\frac{1}{8}\int\Big(\text{u}^{\frac{3}{2}}-\text{u}^{-\frac{1}{2}}\Big)\text{du}$
$\Rightarrow\text{I}=\frac{1}{8}\bigg[\frac{2\text{u}^{\frac{5}{2}}}{5}-\frac{2\text{u}^{\frac{1}{2}}}{1}\bigg]$
$\Rightarrow\text{I}=\frac{1}{8}\Big[\frac{2}{5}\times243-6-\frac{2}{5}\times9\sqrt{3}+2\sqrt{3}\Big]$
$\Rightarrow\text{I}=\frac{1}{8}\Big[\frac{456}{5}-\frac{8\sqrt{3}}{5}\Big]$
$\Rightarrow\text{I}=\frac{57-\sqrt{3}}{5}$
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