Question

Evaluate the integral

\[\int\limits_{}^{}( x/\sqrt{x^2-3x+10})dx\]

Answer 1

Use a trigonometric substitution after completing the square of the expression under the radical.

Answer 2

Write x as m(2x-3)+n
m=1/2 and n=3/2.
so spererating the terms you get
\[(1/2) \int\limits_{}^{}(2x-3)/\sqrt{x ^{2}-3x+10} + \] the second term which is \[3/2\int\limits_{}^{}1/\sqrt{x ^{2}+3x+10}\]

Answer 3

Now both are integrable, the first one is of the form f'(x)/f(x) hence after integration it becomes 2 sqrt(x^2+3x+10). The second term complete the square and write denominator as (x+3/2)^2+10-9/4 And then apply the standard formula.