First slide

Methods of integration

Question

$If \int \frac{\ln (e x^{x + 1}) + {(\ln (x^{\sqrt{x}}))}^{2}}{1 + (x \ln x) (\ln (e^{2} x^{x}))} d x = f (x) + C Where C is arbitrary Constant, then e^{(e^{f (2)} - 1)} is =$

Difficult

A

4
B

2
C

1
D

16

Solution

$\begin{array}{l} Let I = \int \frac{\ln (e x^{x + 1}) + {(\ln (x^{\sqrt{x}}))}^{2}}{1 + (x \ln x) (\ln (e^{2} x^{x}))} d x \\ (\begin{array}{c} Use (1) \log m + \log n = \log m n \\ (2) \log a^{m} = m \log a \end{array}) \\ = \int \frac{\ln (e) + \ln (x^{x + 1}) + (\sqrt{x} \cdot \ln (x))^{2}}{1 + (x \cdot \ln (x)) \cdot (\ln e^{2} + \ln x^{x})} d x \\ = \int \frac{1 + (x + 1) \ln (x) + x (\ln (x))^{2}}{1 + (x \cdot \ln (x)) \cdot (2 + x \cdot \ln x)} d x \\ = \int \frac{1 + \ln x + x \ln x + x (\ln x)^{2}}{1 + 2 (x \ln x) + (x \ln x)^{2}} d x \\ = \int \frac{(1 + x \ln x) (1 + \ln x)}{(1 + x \ln x)^{2}} d x \end{array}$
$\begin{array}{l} = \int \frac{1 + \ln x}{(1 + x \ln x)} d x (∵ \int \frac{f^{'} (x)}{f (x)} = \ln | f (x) | + c) \\ = \ln (1 + x \ln x) + C \\ ∴ f (x) = \ln (1 + x \ln x) \\ \Rightarrow f (2) = \ln | 1 + 2 \ln 2 | \\ \Rightarrow f (2) = \ln | 1 + \ln 4 | \\ \Rightarrow e^{f (2)} = 1 + \ln 4 \\ \Rightarrow e^{f (2)} - 1 = \ln 4 \\ \Rightarrow e^{e^{f (2)} - 1} = 4 \end{array}$
Therefore, the correct answer is (1).

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