use the fundalmental therom of Calculus if applicable Х Evaluate dx x? + 1

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use the fundalmental therom of Calculus if applicable

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Answer 1

Answer #1

To evaluate $\int \frac{x}{\sqrt{x^2+1}}dx$ make the substitution $x=\tan t$ then $dx=\sec^2 tdt$ .

$\int \frac{x}{\sqrt{x^2+1}}dx=\int \frac{\tan t}{\sqrt{\tan^2 t+1}}\sec^2 tdt\\ \int \frac{x}{\sqrt{x^2+1}}dx=\int \frac{\tan t}{\sec t}\sec^2 tdt\\ \int \frac{x}{\sqrt{x^2+1}}dx=\int \tan t\sec tdt\\ \int \frac{x}{\sqrt{x^2+1}}dx=\sec t+C\\ {\color{Blue} \int \frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1}+C}$

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Answer 2

use the fundalmental therom of Calculus if applicable Х Evaluate dx x? + 1

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use the fundalmental therom of Calculus if applicable Х Evaluate dx x? + 1

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