Advertisements
Advertisements
प्रश्न
\[\int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx\]
बेरीज
Advertisements
उत्तर
\[\text{ Let I }= \int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx\]
\[\text{ Also let e}^x \times \frac{1}{x^2} = t \]
\[\text{ Diff both sides w . r . t x }\]
\[ e^x \times \frac{1}{x^2} + e^x \left( \frac{- 2}{x^3} \right) = \frac{dt}{dx}\]
\[ \Rightarrow e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx = dt\]
\[ \therefore \int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx = \int dt\]
\[ = t + C\]
\[ = \frac{e^x}{x^2} + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{x^5 + x^{- 2} + 2}{x^2} dx\]
\[\int \left( \tan x + \cot x \right)^2 dx\]
If f' (x) = x − \[\frac{1}{x^2}\] and f (1) \[\frac{1}{2}, find f(x)\]
\[\int\frac{1 - \cos x}{1 + \cos x} dx\]
` ∫ 1/ {1+ cos 3x} ` dx
\[\int\frac{x^2 + x + 5}{3x + 2} dx\]
\[\int\left( x + 2 \right) \sqrt{3x + 5} \text{dx} \]
\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]
\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x} dx\]
\[\int\frac{1}{1 + \sqrt{x}} dx\]
\[\int\frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} dx\]
` ∫ 1 /{x^{1/3} ( x^{1/3} -1)} ` dx
\[\int\frac{1}{\sin x \cos^3 x} dx\]
\[\int\frac{1}{x^2 + 6x + 13} dx\]
\[\int\frac{e^x}{e^{2x} + 5 e^x + 6} dx\]
\[\int\frac{3 x^5}{1 + x^{12}} dx\]
\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]
\[\int\frac{1}{\cos 2x + 3 \sin^2 x} dx\]
\[\int\frac{1}{1 - 2 \sin x} \text{ dx }\]
\[\int\frac{1}{1 - \sin x + \cos x} \text{ dx }\]
\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx }\]
\[\int\frac{1}{p + q \tan x} \text{ dx }\]
\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]
\[\int\frac{\left( x \tan^{- 1} x \right)}{\left( 1 + x^2 \right)^{3/2}} \text{ dx }\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx }\]
\[\int\frac{1}{x\left( x^n + 1 \right)} dx\]
\[\int\frac{1}{7 + 5 \cos x} dx =\]
\[\int\frac{e^x \left( 1 + x \right)}{\cos^2 \left( x e^x \right)} dx =\]
\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[\int\frac{\sin x}{\sqrt{1 + \sin x}} dx\]
\[\int x \sin^5 x^2 \cos x^2 dx\]
\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx}\]
\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx}\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
