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$\int\left[\frac{(1+\log x)}{\cos ^{2}(x \log x)}\right] d x=$
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Verified Answer
The correct answer is:
$\tan (x \log x)+c$
$I=\int \frac{(1+\log x)}{\cos ^{2}(x \log x)} d x$
Put $x \log x=t \Rightarrow\left[x \cdot \frac{1}{x}+\log x(1)\right] d x=d t \Rightarrow(1-\log x) d x=d t$
$1=\int \frac{1}{\cos ^{2} t} d t=\int \sec ^{2} t d t=\tan t+c$
$=\tan (x \log x)-c$
Put $x \log x=t \Rightarrow\left[x \cdot \frac{1}{x}+\log x(1)\right] d x=d t \Rightarrow(1-\log x) d x=d t$
$1=\int \frac{1}{\cos ^{2} t} d t=\int \sec ^{2} t d t=\tan t+c$
$=\tan (x \log x)-c$
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