Evaluate the given definite integral.
\[
\int_{0}^{\sqrt{3}} \frac{6 x}{\sqrt{x^{2}+1}} d x
\]
\[
\int_{0}^{\sqrt{3}} \frac{6 x}{\sqrt{x^{2}+1}} d x=\text { (Simplify your answer.) }
\]
Answer
The definite integral of the given function from 0 to \( \sqrt{3} \) is \( \boxed{0} \)
Step 1 ：Let \( u = x^{2} + 1 \)
Step 2 ：Then, find the derivative of \( u \) with respect to \( x \), which is \( du/dx = 2x \)
Step 3 ：Substitute these into the integral, we get \( \int_{0}^{\sqrt{3}} 3 du \)
Step 4 ：Solve the integral, we get \( 3u \)
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 \( \boxed{0} \)
