Question

The anti derivative of ƒ(x) = log (log x) + (log x)^–2 whose graph passes through (e, e) is –

• A. x (log (log x) + (log x)^–1)
• B. x (–log (log x) + (log x)^–1 ) + e
• C. x (log (log x) – (log x)^–1) + 2e
• D. None of these

Answer 1

Answer: (C)

x (log (log x) – (log x)^–1) + 2e

Answer 2

An antiderivative of ƒ(x) = F(x) =

$\int_{}^{}{(\log(\log x) + (\log x)^{–2})}$dx + c

= x log (log x) – $\int_{}^{}\frac{x}{x\log x}$dx+ $\int_{}^{}{(\log x)^{- 2}}$dx + c

(integrating by parts the first term log x is)

= x log (log x) – [x (log x)^–1 + $\int_{}^{}{(\log x)^{- 2}}$dx]

+ $\int_{}^{}{(\log x)^{- 2}}$dx + c

(again integrating by parts (log x)^–1 as first term)

= x log (log x) – x (log x)^–1 + c

Putting x = e, we have e = 0 – e + c so c = 2e

Thus F (x) = x (log (log x) – (log x)^–1 + 2e.