By the chain rule, we have ( x 2 − 3 ) ′ = 1 2 x 2 − 3 ⋅ ( x 2 − 3 ) ′ = x x 2 − 3 {\displaystyle ({\sqrt {x^{2}-3}})^{\prime }={\frac {1}{2{\sqrt {x^{2}-3}}}}\cdot (x^{2}-3)'={\frac {x}{\sqrt {x^{2}-3}}}} .
Then, by the definition of anti-derivative, we get
∫ 3 x x 2 − 3 = 3 ∫ x x 2 − 3 = 3 ∫ ( x 2 − 3 ) ′ = 3 x 2 − 3 + C {\displaystyle \int {\frac {3x}{\sqrt {x^{2}-3}}}=3\int {\frac {x}{\sqrt {x^{2}-3}}}=3\int ({\sqrt {x^{2}-3}})^{\prime }=3{\sqrt {x^{2}-3}}+C} .
Answer: 3 x 2 − 3 + C {\displaystyle \color {blue}{3{\sqrt {x^{2}-3}}}+C}
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