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