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\( \int \frac{1}{\sqrt{x}+x \sqrt{x}} d x= \)
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Verified Answer
The correct answer is:
\( 2 \tan ^{-1} \sqrt{x}+C \)
(B)
\( I=\int \frac{1}{\sqrt{x}\left[1+(\sqrt{x})^{2}\right]} d x \)
Put \( \sqrt{x}=t \Rightarrow \frac{1}{\sqrt{x}} d x=2 d t \)
\( \therefore I=\int \frac{2 d t}{1+t^{2}}=2 \tan ^{-1} t+C=2 \tan ^{-1} \sqrt{x}+C \)
\( I=\int \frac{1}{\sqrt{x}\left[1+(\sqrt{x})^{2}\right]} d x \)
Put \( \sqrt{x}=t \Rightarrow \frac{1}{\sqrt{x}} d x=2 d t \)
\( \therefore I=\int \frac{2 d t}{1+t^{2}}=2 \tan ^{-1} t+C=2 \tan ^{-1} \sqrt{x}+C \)
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