MHB Find Velocity of Particles: Indefinite Integrals

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Finding the velocity of particles involves evaluating the indefinite integral of the acceleration function, expressed as v = ∫ a(t) dt. The discussion highlights the application of integration by parts (IBP) to derive the integral of the function 125t^4ln^2(t). The formula derived through this method is v = 25t^5ln^2(t) - 10t^5ln(t) + 2t^5 + C. Participants express willingness to assist with further derivations and provide additional resources for understanding IBP. The focus remains on the mathematical techniques necessary for solving these integrals.
Madu
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To help find the velocity of particles requires the evaluation of the indefinite integral of the acceleration
function, a(t), i.e.
v = Z a(t) dt.

Your help greatly appreciated.
 

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I would think IBP could be used to obtain:

$$\int 125t^4\ln^2(t)\,dt=t^5\left(25\ln^2(t)-10\ln(t)+2\right)+C$$

If you want help actually deriving this formula, please let me know. :)
 
For some useful information in understanding IBP, see here.

$$\begin{align*}\int 125t^4\ln^2(t)\,dt&=25t^5\ln^2(t)-50\int t^4\ln(t)\,dt \\
&=25t^5\ln^2(t)-50\left(\frac15t^5\ln(t)-\frac15\int t^4\,dt\right) \\
&=25t^5\ln^2(t)-10t^5\ln(t)+2t^5+C\end{align*}$$
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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