Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Solution 1

Consider the substitution $u=x^{2}-3$. Then $du=2x\,dx$ because the derivative of $x^{2}-3$ with respect to $x$ is $2x$. Hence,

$\int {\dfrac {3x}{\sqrt {x^{2}-3}}}\,dx={\dfrac {3}{2}}\int {\dfrac {2x\,dx}{\sqrt {x^{2}-3}}}={\dfrac {3}{2}}\int {\dfrac {du}{\sqrt {u}}}$

By the Power Rule for the integration, we get

$\int {\dfrac {du}{\sqrt {u}}}=\int u^{-{\frac {1}{2}}}du={\frac {1}{-{\frac {1}{2}}+1}}u^{-{\frac {1}{2}}+1}+C=2{\sqrt {u}}+C$

where $C$ is an arbitrary constant (the constant of integration)

So,

$\int {\dfrac {3x}{\sqrt {x^{2}-3}}}\,dx={\dfrac {3}{2}}\int {\dfrac {du}{\sqrt {u}}}={\dfrac {3}{2}}(2{\sqrt {u}}+C)$

and substituting $x^{2}-3$ for $u$ into the above expression now gives

$\int {\dfrac {3x}{\sqrt {x^{2}-3}}}\,dx=3{\sqrt {u}}+C=\color {blue}3{\sqrt {x^{2}-3}}+C$