Question

integration of (1/2-3x)+1/√(3x-2) dx

Answer 1

Answer:

dx1+3x2

Substitute:

x=13√tanu

dx=13√sec2udu

∫13√sec2udu1+3(13√tanu)2

Pull out that constant:

13–√∫sec2udu1+3(13tan2u)

Do the math:

13–√∫sec2udu1+tan2u

Trigonometric identity: 1+tan2u=sec2u:

13–√∫sec2udusec2u

Cancel:

13–√∫du

That looks pretty easy:

13–√u+C

Undo the substitution:

13–√arctan(3–√x)+C

Check the answer on desmos: