First slide

Methods of integration

Question

$If \int \frac{1}{x {((\log x)^{2} - 2 \log x + 10)}^{2}} d x = \frac{1}{54} [\tan^{- 1} (f (x)) + \frac{3 (\log x - 1)}{g (x)}] + C then f (e) + g (e) = - - -$

Moderate

A

9
B

1
C

0
D

None

Solution

$\begin{array}{l} \int \frac{1}{x {((\log x)^{2} - 2 \log x + 10)}^{2}} d x \\ \log x = t \Rightarrow \frac{1}{x} d x = d t \\ I = \int \frac{1}{{((t - 1)^{2} + 9)}^{2}} d t \\ Put t - 1 = 3 \tan θ \\ d t = 3 \sec^{2} θ d θ \\ I = \int \frac{3 \sec^{2} θ}{{(9 \sec^{2} θ)}^{2}} d θ \\ = \frac{1}{27} \int \cos^{2} θ d θ \\ = \frac{1}{27} \int \frac{1}{2} + \frac{1}{2} \cos 2 θ d θ \end{array}$
$\begin{array}{l} = \frac{1}{54} [θ + \frac{\sin 2 θ}{2}] + C \\ = \frac{1}{54} [\tan^{- 1} (\frac{t - 1}{3}) + \frac{1}{2} \frac{2 \tan θ}{1 + \tan^{2} θ}] + C \end{array}$
$\begin{array}{l} = \frac{1}{54} [\tan^{- 1} (\frac{t - 1}{3}) + \frac{3 (t - 1)}{t^{2} - 2 t + 10}] + C \\ f (x) = \frac{\log x - 1}{3}, g (x) = (\log x)^{2} - 2 \log x + 10 \end{array}$

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