Advertisements
Advertisements
Question
Solution
\[\text{ Let I }= \int e^x \left( \frac{x - 1}{2 x^2} \right)dx\]
\[ = \frac{1}{2}\int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right)dx\]
\[\text{ here }\frac{1}{x} = f(x) \text{ Put e}^x f(x) = t\]
\[ \Rightarrow - \frac{1}{x^2} = f'(x)\]
\[\text{ let e}^x \frac{1}{x} = t\]
\[\text{ Diff both sides w . r . t x}\]
\[\left( e^x \frac{1}{x} + e^x \frac{- 1}{x^2} \right) = \frac{dt}{dx}\]
\[ \Rightarrow e^x \left( \frac{1}{x} - \frac{1}{x^2} \right)dx = dt\]
\[ \therefore I = \frac{1}{2}\int dt\]
\[ = \frac{t}{2} + C\]
\[ = \frac{e^x}{2x} + C\]
APPEARS IN
RELATED QUESTIONS
\[\int\sqrt{x}\left( 3 - 5x \right) dx\]
If f' (x) = x − \[\frac{1}{x^2}\] and f (1) \[\frac{1}{2}, find f(x)\]
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
\[\int \tan^2 \left( 2x - 3 \right) dx\]
` = ∫ root (3){ cos^2 x} sin x dx `
` ∫ 1 /{x^{1/3} ( x^{1/3} -1)} ` dx
\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]