Exercise is mixed —some require integration by parts, while others can be integrated by using techniques discussed in the chapter on integration.
∫(2x-1) ln (3x) dx.

ln(3x)-x²/2 + x +c

Step-by-step explanation:

∫(2x-1) ln (3x) dx can be computed with parts. The ln part can be made simpler be derivating, and integrating a polynomial wont hurt us that much.

We can derivate ln(3x) using the chain rule, the derivate of ln(x) is 1/x and the derivate of 3x is 3; therefore

(ln(3x))' = (1/3x)*3 = 1/x

A primitive of (2x-1) is, on the other hand, x²-x

Hence

∫(2x-1) = ln(3x) (x²-x) - ∫(x²-x)/x dx = ln(3x) (x²-x) - ∫(x-1) dx = ln(3x) (x²-x) - (x²/2 - x + k) = ln(3x)-x²/2 + x +c