For x - 0 - let f x - -1xlog t-t-1dt- The value of 98 f e -f 1-e is

f( x ) =∫ _ 1 ^ x log nbsp;t t+1 dt ⇒ f( 1 x nbsp;) =∫ _ 1 ^ x log nbsp;t t( t+1 ) nbsp; dt ∴ f( x ) +f( 1 x nbsp;) =∫ _ 1 ^ x ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Integration Using Substitution

For x > 0 ,...

Question

For $x > 0$ , let $f (x) = \int_{1}^{x} \frac{log t}{t + 1} d t$ . The value of $98 (f (e) + f (1 / e))$ is

Open in App

Solution

Suggest Corrections

Similar questions

F (x) = \int_{1}^{x} \frac{log t}{1 + t + t^{2}} d t

F (x) = - F (1 / x)

F (x) = \int_{1}^{x} \frac{log t}{t + 1} d t

F (x) + F (1 / x) = (1 / 2) (log x)^{2}

f (x) = \int_{1}^{x} \frac{log t}{t + 1} d t

f (x) + f (1 / x) = k {(log x)}^{2}

k

For x > 0, let f (x) = x \int 1 \frac{log t}{1 + t} d t . Then f (x) + f (\frac{1}{x}) is equal to

x > 0

f (x) = \int_{1}^{x} \frac{log t}{1 + t} d t

f (x) + f (\frac{1}{x})

For $x > 0$ , let $f (x) = \int_{1}^{x} \frac{log t}{1 + t} d t$ . Then $f (x) + f (\frac{1}{x})$ is equal to

Join BYJU'S Learning Program