Advertisements
Advertisements
Question
Advertisements
Solution
\[\text{ Let I }= \int e^x \left( \frac{x - 4}{\left( x - 2 \right)^3} \right)dx\]
\[ = \int e^x \left[ \frac{x - 2 - 2}{\left( x - 2 \right)^3} \right]dx\]
\[ = \int e^x \left[ \frac{1}{\left( x - 2 \right)^2} - \frac{2}{\left( x - 2 \right)^3} \right]dx\]
\[\text{ Here, f(x) }= \frac{1}{\left( x - 2 \right)^2}\]
\[ \Rightarrow f'(x) = \frac{- 2}{\left( x - 2 \right)^3}\]
\[\text{ Put e}^x f(x) = t\]
\[ \Rightarrow e^x \frac{1}{\left( x - 2 \right)^2} = t\]
\[\text{ Diff both sides w . r . t x }\]
\[\left[ e^x \frac{1}{\left( x - 2 \right)^2} + e^x \frac{- 2}{\left( x - 2 \right)^3} \right]dx = dt\]
\[ \text{∴ I }= \int dt\]
\[ = t + C\]
\[ = \frac{e^x}{\left( x - 2 \right)^2} + C\]
APPEARS IN
RELATED QUESTIONS
` = ∫1/{sin^3 x cos^ 2x} dx`
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
