Evaluate the given definite integral.
$\int_{0}^{\sqrt{3}} \frac{6 x}{\sqrt{x^{2} + 1}} d x$
$\int_{0}^{\sqrt{3}} \frac{6 x}{\sqrt{x^{2} + 1}} d x = (Simplify your answer.)$

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Answer

The definite integral of the given function from 0 to $\sqrt{3}$ is $0$

Steps

Step 1 ：Let $u = x^{2} + 1$

Step 2 ：Then, find the derivative of $u$ with respect to $x$ , which is $d u / d x = 2 x$

Step 3 ：Substitute these into the integral, we get $\int_{0}^{\sqrt{3}} 3 d u$

Step 4 ：Solve the integral, we get $3 u$

Step 5 ：Substitute $u$ back into the original variable $x$ , we get $3 (x^{2} + 1)$

Step 6 ：Evaluate the definite integral from the given limits of integration, we get $3 ({\sqrt{3}}^{2} + 1) - 3 (0^{2} + 1) = 3 \sqrt{3} - 3$

Step 7 ：The definite integral of the given function from 0 to $\sqrt{3}$ is $0$