Karnataka Board PUCPUC Science 2nd PUC Class 12

I F ∫ ( X − 1 X 2 ) E X D X = F ( X ) E X + C , T H E N W R I T E T H E V a L U E O F F ( X ) . - Mathematics

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Sum
\[\text{ If } \int\left( \frac{x - 1}{x^2} \right) e^x dx = f\left( x \right) e^x + C, \text{ then  write  the value of  f}\left( x \right) .\]
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Solution

\[\int\left( \frac{x - 1}{x^2} \right) e^x dx = \int\left( \frac{x}{x^2} - \frac{1}{x^2} \right) e^x dx\]
\[ = \int\left( \frac{1}{x} - \frac{1}{x^2} \right) e^x dx\]
\[\text{ Consider,} f\left( x \right) = \frac{1}{x},\text{  then f}^ \left( x \right) = - \frac{1}{x^2}\]
\[\text{ Thus , the  given  integrand  is  of  the form e}^x \left[ f\left( x \right) + f^ \left( x \right) \right] . \]
\[\text{ Therefore, }\int\left( \frac{x - 1}{x^2} \right) e^x dx = \frac{1}{x} e^x + C\]
\[\text{ Hence,} f\left( x \right) = \frac{1}{x} .\]

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Chapter 19: Indefinite Integrals - Very Short Answers [Page 198]

APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Very Short Answers | Q 56 | Page 198

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