Question
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{x}}(\text{x}-4)}{(\text{x}-2)^3}\text{dx}$
$\int\frac{\text{e}^{\text{x}}(\text{x}-4)}{(\text{x}-2)^3}\text{dx}$
✓
Answer
Let $\text{I}=\int\frac{\text{e}^{\text{x}}(\text{x}-4)}{(\text{x}-2)^3}\text{dx}$
$=\int\text{e}^{\text{x}}\bigg\{\frac{(\text{x}-2)-2}{(\text{x}-2)^3}\bigg\}\text{dx}$
$=\int\text{e}^{\text{x}}\bigg\{\frac{1}{(\text{x}-2)^2}-\frac{2}{(\text{x}-2)^3}\bigg\}\text{dx}$
Here, $\text{f(x)}=\frac{1}{(\text{x}-2)^2}$ and $\text{f}'\text{(x)}=\frac{-2}{(\text{x}-2)^3}$
And we know that,
$\int\text{e}^{\text{ax}}(\text{af(x)}+\text{f}'(\text{x}))\text{dx}=\text{e}^{\text{ax}}\text{f(x)}+\text{C}$
$\therefore\int\text{e}^{\text{x}}\bigg\{\frac{1}{(\text{x}-2)^2}-\frac{2}{(\text{x}-2)^3}\bigg\}\text{dx}=\frac{\text{e}^{\text{x}}}{(\text{x}-2)^2}+\text{C}$
$\therefore\text{I}=\frac{\text{e}^{\text{x}}}{(\text{x}-2)^2}+\text{C}$
$=\int\text{e}^{\text{x}}\bigg\{\frac{(\text{x}-2)-2}{(\text{x}-2)^3}\bigg\}\text{dx}$
$=\int\text{e}^{\text{x}}\bigg\{\frac{1}{(\text{x}-2)^2}-\frac{2}{(\text{x}-2)^3}\bigg\}\text{dx}$
Here, $\text{f(x)}=\frac{1}{(\text{x}-2)^2}$ and $\text{f}'\text{(x)}=\frac{-2}{(\text{x}-2)^3}$
And we know that,
$\int\text{e}^{\text{ax}}(\text{af(x)}+\text{f}'(\text{x}))\text{dx}=\text{e}^{\text{ax}}\text{f(x)}+\text{C}$
$\therefore\int\text{e}^{\text{x}}\bigg\{\frac{1}{(\text{x}-2)^2}-\frac{2}{(\text{x}-2)^3}\bigg\}\text{dx}=\frac{\text{e}^{\text{x}}}{(\text{x}-2)^2}+\text{C}$
$\therefore\text{I}=\frac{\text{e}^{\text{x}}}{(\text{x}-2)^2}+\text{C}$
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