integral 5x16x29 using inverse like arcsin please show steps

Solution

(5x - 6)/(x^2 + 9) dx= 5x/(x^2 + 9) - 6/(x^2 + 9) dx= 5 x/(x^2 + 9) dx - 6 1/(x^2 + 9) dx= 5/2 2x/(x^2 + 9) dx - 6 1/(x^2 + 9) dx
For the first integral, let u = x^2 + 9 <==> du = 2x dx to get:
5/2 1/u du= 5ln|u|/2 + C= 5ln(x^2 + 9)/2 + C (as x^2 + 9 > 0 for all x)
Then:
5/2 2x/(x^2 + 9) dx - 6 1/(x^2 + 9) dx= 5ln(x^2 + 9)/2 - 6 1/(x^2 + 9) dx
Since the integral of 1/(x^a + a^2) is (1/a)arctan(x/a) + C, we can conclude with:
5ln(x^2 + 9)/2 - 3arctan(x/3) + C <== ANSWER

i have used arctan..i hope you can convert arctan to arcsin

this should help you