1. -x cos(x) + sin(x) + c

1. ln2(x) / 2 + c

   1. We do know how to integrate both ln(x) and 1/x. Let’s think about the choice for u: if u = 1/x, then du = x-2, which will be worse. Let’s take u = ln(x), dv = 1/x, so du = 1/x, v = ln(x) Looks funny but it works: ∫ ln(x) / x dx = ln2(x) – ∫ ln(x) / x dx On the right hand side, we have thesame integral that we started with – with the opposite sign. Gather them up on the left and we get:

∫ ln(x) / x dx = ln2(x) / 2 + c

1. -ln(x) – 1/x + c

1. [ x² – 2x + 2 ] e^x

The easiest choice for variables is: u = x², dv = e^x, so du = 2x, v = e^x ∫ x² e^x dx =x² e^x – ∫ 2x e^x dx We could integrate the right hand side by parts, but we have already done that in the notes so we can use the answer, which said: ∫ x e^x dx = x e^x – e^x Altogether then, we have ∫ x² e^x dx = [x² – 2x + 2] e^x

1. x arctan(x) -1/2 ln [1 + x²]

1. We have no idea how to integrate arctan(x) (which is the point here), so we cannot make it equal to dv. That leaves us with: u = arctan(x), dv = 1, so du = 1/(1 + x²), v = x Now we get: ∫arctan(x) dx = x arctan(x) – ∫ x/(1 + x²) dx = x arctan(x) -1/2 ln [1 + x²]