Advertisements
Advertisements
प्रश्न
उत्तर
\[\text{Let I} = \int\frac{e^{2x}}{e^{2x} - 2}dx\]
\[\text{Putting }e^{2x} = t\]
\[ \Rightarrow 2 e^{2x} = \frac{dt}{dx}\]
\[ \Rightarrow e^{2x} dx = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int\frac{1}{t - 2}dt\]
\[ = \frac{1}{2} \text{ln }\left| t - 2 \right| + C\]
\[ = \frac{1}{2} \text{ln }\left| e^{2x} - 2 \right| + C \left[ \because t = e^{2x} \right]\]
APPEARS IN
संबंधित प्रश्न
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
Evaluate the following integrals:
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
\[\int\frac{\cot x}{\sqrt{\sin x}} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integrals:
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int sqrt(5 - 2x + x^2) "d"x`
Evaluate the following:
`int x/(x^4 - 1) "d"x`
Evaluate the following:
`int sqrt(2"a"x - x^2) "d"x`
Evaluate the following:
`int sqrt(x)/(sqrt("a"^3 - x^3)) "d"x`