Advertisement Remove all ads

∫ ( 1 Log X − 1 ( Log X ) 2 ) D X - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum
\[\int\left( \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right) dx\]
Advertisement Remove all ads

Solution

\[\text{  Let I  }= \int\left( \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right)dx\]

\[\text{ Put  log x }= t\]

\[ \Rightarrow x = e^t \]

\[ \Rightarrow dx = e^t dt\]

\[ \therefore I = \int e^t \left( \frac{1}{t} - \frac{1}{t^2} \right)dt\]

\[\text{ Here}, f(t) = \frac{1}{t}\]

\[ \Rightarrow f'(t) = \frac{- 1}{t^2}\]

\[\text{ let e} ^t \times \frac{1}{t} = p\]

\[\text{ Diff  both  sides  w . r . t  t}\]

\[\left( e^t \times \frac{1}{t} + e^t \times \frac{- 1}{t^2} \right)dt = dp\]

\[ \therefore I = \int dp\]

\[ = p + C\]

\[ = \frac{e^t}{t} + C\]

\[ = \frac{x}{\log x} + C\]

Concept: Indefinite Integral Problems
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.26 | Q 20 | Page 143

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×