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Indefinite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integration4 Marks
Question
Evaluate : $\int\left[\log (\log x)+\frac{1}{(\log x)^2}\right] \cdot d x $
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Answer

$\begin{aligned}\text {I } \quad=\int \log (\log x) \cdot 1 \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot \int 1 \cdot d x-\int \frac{d}{d x} \cdot \log (\log x) \int 1 \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot x-\int \frac{1}{\log x} \cdot \frac{1}{x} \cdot(x) \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot x-\int \frac{1}{\log x} \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot x-\int(\log x)^{-1} \cdot 1 \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot x-\left\{(\log x)^{-1} \cdot \int 1 \cdot d x+\int \frac{d}{d x} \cdot(\log x)^{-1} \cdot \int 1 \cdot d x \cdot d x\right\}+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot x-\left\{(\log x)^{-1} \cdot x-\int-1(\log x)^{-2} \cdot \frac{1}{x} \cdot x \cdot d x\right\}+\int \frac{1}{(\log x)^2} \cdot d x \\ & =\log (\log x) \cdot x-(\log x)^{-1} \cdot x-\int(\log x)^{-2} \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & =x \cdot \log (\log x)-\frac{x}{\log x}-\int \frac{1}{(\log x)^2} \cdot d x+\int \frac{1}{(\log x)^2} \cdot d x \\ & \therefore \int\left[\log (\log x)+\frac{1}{(\log x)^2}\right] \cdot d x=x \cdot \log (\log x)-\frac{x}{\log x}+c \\ & \end{aligned}$

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