Question
Evaluate the following integrals:
$\int \frac{\text{x}^2}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
$\int \frac{\text{x}^2}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
✓
Answer
We have,
$\text{I}=\int \frac{\text{x}^2}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
Putting $\text{x}+2=\text{t}^2$
$\text{x}=\text{t}^2-2$
Diff both sides
$\text{dx}=2\text{t dt} $
$\text{I}=\int\frac{(\text{t}^2-2)^2}{(\text{t}^2-2-1)\text{t}}2\text{t dt}$
$=2\int\frac{(\text{t}^2-2)^2\text{dt}}{\text{t}^2-3}$
$=2\int\frac{(\text{t}^4-4\text{t}^2+4)}{\text{t}^2-3}\text{ dt}$
Dividing numerator by denominator, we get
$\therefore\ \text{I}=2\int\Big(\text{t}^2-1+\frac{1}{\text{t}^2-3}\Big)\text{ dt}$
$=2\int\text{t}^2\text{ dt}-2\int\text{ dt}+2\int\frac{\text{dt}}{\text{t}^2-(\sqrt{3})^2}$
$=2\Big[\frac{\text{t}^3}{3}\Big]-2\text{t}+2\times\frac{1}{2\sqrt{3}}\log\Big|\frac{\text{t}-\sqrt{3}}{\text{t}+\sqrt{3}}\Big|+\text{C}$
$=\frac{2}{3}(\sqrt{\text{x}+2})^3-2\sqrt{\text{x}+2}+\frac{1}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}+2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}$
$=\frac{2}{3}(\text{x}+2)^{\frac{3}{2}}-2\sqrt{\text{x}+2}+\frac{1}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}+2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}$
$\text{I}=\int \frac{\text{x}^2}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
Putting $\text{x}+2=\text{t}^2$
$\text{x}=\text{t}^2-2$
Diff both sides
$\text{dx}=2\text{t dt} $
$\text{I}=\int\frac{(\text{t}^2-2)^2}{(\text{t}^2-2-1)\text{t}}2\text{t dt}$
$=2\int\frac{(\text{t}^2-2)^2\text{dt}}{\text{t}^2-3}$
$=2\int\frac{(\text{t}^4-4\text{t}^2+4)}{\text{t}^2-3}\text{ dt}$
Dividing numerator by denominator, we get
$\therefore\ \text{I}=2\int\Big(\text{t}^2-1+\frac{1}{\text{t}^2-3}\Big)\text{ dt}$
$=2\int\text{t}^2\text{ dt}-2\int\text{ dt}+2\int\frac{\text{dt}}{\text{t}^2-(\sqrt{3})^2}$
$=2\Big[\frac{\text{t}^3}{3}\Big]-2\text{t}+2\times\frac{1}{2\sqrt{3}}\log\Big|\frac{\text{t}-\sqrt{3}}{\text{t}+\sqrt{3}}\Big|+\text{C}$
$=\frac{2}{3}(\sqrt{\text{x}+2})^3-2\sqrt{\text{x}+2}+\frac{1}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}+2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}$
$=\frac{2}{3}(\text{x}+2)^{\frac{3}{2}}-2\sqrt{\text{x}+2}+\frac{1}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}+2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}$
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