1. x ln²(x) – 2x ln(x) + 2x + c

1. Recall that ∫ln(x) dx= x ln(x) – x. We can either take dv = 1 or dv = ln(x), since we know how to integrate both. The former is perhaps slightly easier. We have: dv= 1, u= ln²(x), so v= x, du= 2 ln(x) / x Now we integrate by parts: ∫ln²(x) dx= x ln²(x) – ∫2ln(x) dx = x ln²(x) – 2x ln(x) + 2x + c You can try starting with dv= ln(x) for practice, see if you can get the same answer.

1. 1/2 [sin(x) cos(x) + x + c]

1. Use u= dv= cos(x). Then du= – sin(x), v= sin(x) and we get ∫cos²(x) dx= sin(x) cos(x) + ∫sin²(x) dx Now we substitute [1 – cos²(x)] for the sin²(x) and the integral on the right side becomes ∫sin²(x) dx= ∫[1 – cos²(x)] dx = x – ∫cos²(x) dx We put this into the original equation, and gather up both cos2(x) terms to get ∫cos²(x) dx = 1/2 [sin(x) cos(x) + x + c]

1. x/2 (cos[ln(x)] + sin[ln(x)]) + c

1. The only choice here is to set dv = 1: u = cos[ ln(x) ], dv = 1, so v = x and du = – sin[ ln(x) ] / x Integrating by parts: ∫ cos[ ln(x) ] dx = x cos[ ln(x) ] + ∫ sin[ ln(x) ] dx This does not seem to be of much help. We can try integrating ∫sin[ln(x)] dx by parts, to see if we can get to anything useful. Again, set dv = 1: u = sin[ln(x)], dv = 1, so v = x and du = cos[ln(x)]/x We get: ∫sin[ln(x)] dx = x sin[ln(x)] – ∫cos[ ln(x)] dx Altogether, we have ∫cos[ln(x)] dx = x cos[ln(x)] + x sin[ln(x)] – ∫cos[ ln(x)] dx We can put both integrals on the left, and arrive at the answer: ∫cos[ln(x)] dx = x/2 ( cos[ ln(x) ] + sin[ ln(x) ] ) + c

1. 1/2 [e^x sin(x) + e^x cos(x)] + c

1. In this case, it is not at all clear which factor we should pick for u and which for dv. It turns out that the choices lead to a very similar solution. Let us choose dv = cos(x). u = ex, dv = cos(x), so du = ex, v = sin(x) Integrating by parts: ∫e^xcos(x) dx= e^x sin(x) – ∫e^x sin(x) dx We integrate by parts again, choosing u= e^x again (otherwise we would get back where we started): u = e^x, dv= sin(x), so du= ex, v= – cos(x) We get: ∫e^x sin(x) dx = -e^x cos(x) + ∫e^x cos(x) dx Altogether, we have: ∫e^x cos(x) dx= e^x sin(x) + e^x cos(x) – ∫e^x cos(x) dx Combining the integrals gives us the answer: ∫e^{x cos(x) dx = 1/2 [ex sin(x) + ex cos(x)] + c}

1. 1/101 x (x-1)¹⁰¹ – 1/101 ⋅ 1/102 (x-1)¹⁰² + c

1. Let’s not multiply it out. Instead, use integration by parts, with u = x: u = x, dv = (x-1)¹⁰⁰, so du = 1, v = 1/101 (x-1)¹⁰¹ We get: ∫x (x-1)¹⁰⁰ dx = 1/101 x(x-1)¹⁰¹ – ∫1/101 (x-1)¹⁰¹ dx = 1/101 x (x-1)¹⁰¹ – 1/101 ⋅ 1/102 (x-1)¹⁰² + c