t ln(t + 5) dt find indefinite integral

Question

indefinite integral

Answer 1

By parts: u = ln(t + 5: dv = tdt. Then du = 1/(t + 5)dt; v = t²/2

.∫tln(t + 5)dt = t²/2ln(t + 5) - 1/2∫t²/(t + 5)dt = t²ln(t + 5)/2 - 1/2∫[t - 5 + 25/(t + 5)]dt = t²ln(t + 5)/2 - t²/4 + 5t/2 -

25ln(t + 5)/2 + C. Here t + 5 > 0

Answer 2

There are four methods of integration you should know. U-sub, trig-sub, partial fractions, and by-parts. There are no real signs to use the first three methods so you can try by-parts. You should chose u and dv by what you know the derivative and antiderivative of. That makes u = ln(t+5) and dv = t dt. Using

∫ u dv = u v - ∫ v du,

∫ t/ln(t+5) dt = 1/2 t² ln(t+5) - ∫ 1/2 t²/(t+5) dt.

In integrals (math in general, life) you always want to turn something you don't know how to do into something you have done before. The last integral is from the u-sub section so it looks like you can do it. Using polynomial division (synthetic division),

1/2 t²/(t+5) = t/2 + 25/2 1/(t+5) - 5/2.

Then -∫ 1/2 t²/(t+5) dt = -t²/4 - 25/2 ln(t+5) + 5/2 t + C

Combing these two results should give you the answer. You can check your work in Wolfram-Alpha.

https://www.wolframalpha.com/input?i=int%28t*ln%28t%2B5%29%29