∫( 1/(x？ -x-6))dx =( 1/5)∫( 1/(x-3)- 1/(x+2))dx =( 1/5)(∫( 1/(x-3))d(x-3)-( 1/(x+2))d(x+2))=( 1/5)(ln | x-3 |-ln

∫cosxe^(sinx)dx=∫e^(sinx)d(sinx)=e^(sinx)+c

∫ (dx)/(√ x) (1+x) order √ x = t, then x = t? So the original formula = ∫ d (t? )/(t( 1+t？ ))=2∫( 1/( 1+t？ )) dt = 2arctant, t = √ x, so the answer is = 2arctan √ x+c.

∫( 1/( 1+? √x))dx makes t =? √x So x = t? So there is the original formula = ∫ d (t? )/( 1+t)=3∫t？ /( 1+t)dt=3∫(t？ - 1+ 1)/∫∫( 1+t)dt = 3∫((t？ - 1)/( 1+t)+ 1/( 1+t))dt = 3∫(t- 1)+( 1/(t+ 1))dt = 3(( 1/2)(t- 1)？ +ln | t+ 1 |)+c and t =? X So the answer is (3/2) (? √x- 1)？ +ln|？ √x+ 1|)+C

∫? √ x/(√ x+ 1) dx order t? = x is the original formula = ∫ t/(t? + 1)d(t？ )=4∫t？ /(t？ + 1)dt=4∫(t？ - 1+ 1)/(t？ + 1)dt=4∫((t？ - 1)+ 1/(t？ + 1))dt=4(( 1/3)t？ -t)+4arctant+c and t =? √x So the answer = 4 (( 1/3) (? √x)？ -? √x)+4arctan(？ √x)+C