First slide

Methods of integration

Question

The antiderivative of $f (x) = \log (\log x) +$ $(\log x)^{- 2}$
whose graph passes through $(e, e)$ is

Moderate

A

$x (\log (\log x) + (\log x)^{- 1})$
B

$x (- \log (\log x) + (\log x)^{- 1}) + e$
C

$x (\log (\log x) - (\log x)^{- 1}) + 2 e$
D

none of these

Solution

An antiderivative of $f (x) = F (x) =$
$\begin{matrix} \int (\log (\log x) + (\log x)^{- 2}) d x + C \\ = x \log (\log x) - \int \frac{x}{x \log x} d x + \int (\log x)^{- 2} d x + C \end{matrix}$
$(integrating by parts the first term)$
$\begin{array}{l} = x \log (\log x) - [x (\log x)^{- 1} + \int (\log x)^{- 2} d x] + \\ \int (\log x)^{- 2} d x + C \end{array}$
$\begin{aligned} (again integrating by parts (\log x)^{- 1}) \\ = x \log (\log x) - x (\log x)^{- 1} + C \end{aligned}$
Putting $x = e$ we have $e = 0 - e + C so C = 2 e$
Thus $F (x) = x (\log (\log x) - (\log x)^{- 1}) + 2 e$

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