Evaluate the following integrals: x (x-3) x+1 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}}{(\text{x}-3)\sqrt{\text{x}+1}}\text{ dx}$

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Answer

Let $\text{I}=\int\frac{\text{x}}{(\text{x}-3)\sqrt{\text{x}+1}}\text{ dx}$
$=\int\frac{(\text{x}-3)+3}{(\text{x}-3)\sqrt{\text{x}+1}}\text{ dx}$
$\text{I}=\int\frac{\text{dx}}{\sqrt{\text{x}+1}}+3\int\frac{\text{dx}}{(\text{x}-3)\sqrt{\text{x}+1}}\ ...(\text{i})$
Now, $\int\frac{\text{dx}}{\sqrt{\text{x}+1}}=2\sqrt{\text{x}+1}+\text{C}_1$
and, $\int\frac{\text{dx}}{\sqrt{\text{x}+2}}=2\sqrt{\text{x}+2}+\text{C}_2$
$\int\frac{\text{dx}}{(\text{x}-3)\sqrt{\text{x}+1}}$
Let $\text{x}+1=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\int\frac{\text{dx}}{(\text{x})-3\sqrt{\text{x}+1}}=2\int\frac{\text{t dt}}{(\text{t}^2-4)\text{t}}$
$=2\Big|\frac{\text{dt}}{\text{t}^2-4}\Big|$
$=\frac{2}{2\times2}\log\Big|\frac{\text{t}-2}{\text{t}+2}\Big|+\text{C}_2$
$\therefore\ \int\frac{\text{dx}}{(\text{x}-3)\sqrt{\text{x}+1}}=\frac{1}{2}\log\bigg|\frac{\sqrt{\text{x}+1}-2}{\sqrt{\text{x}}+1+2}\bigg|+\text{C}_2$
Thus, from (i)
$\text{I}=2\sqrt{\text{x}+1}+\frac{3}{2}\log\bigg|\frac{\sqrt{\text{x}+1}-2}{\sqrt{\text{x}+1}+2}\bigg|+\text{C}$ [When C = C₁ + C₂]

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