Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{ Let I }= \int\left( \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right)dx\]
\[\text{ Put log x }= t\]
\[ \Rightarrow x = e^t \]
\[ \Rightarrow dx = e^t dt\]
\[ \therefore I = \int e^t \left( \frac{1}{t} - \frac{1}{t^2} \right)dt\]
\[\text{ Here}, f(t) = \frac{1}{t}\]
\[ \Rightarrow f'(t) = \frac{- 1}{t^2}\]
\[\text{ let e} ^t \times \frac{1}{t} = p\]
\[\text{ Diff both sides w . r . t t}\]
\[\left( e^t \times \frac{1}{t} + e^t \times \frac{- 1}{t^2} \right)dt = dp\]
\[ \therefore I = \int dp\]
\[ = p + C\]
\[ = \frac{e^t}{t} + C\]
\[ = \frac{x}{\log x} + C\]
APPEARS IN
संबंधित प्रश्न
\[\int\left\{ x^2 + e^{\log x}+ \left( \frac{e}{2} \right)^x \right\} dx\]
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
Write a value of
If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]
\[\int \sin^3 \left( 2x + 1 \right) \text{dx}\]
