What is the integral of (ln t)^2?

Question

Just remember the squared exponent is around the whole function, not just on the t.

Answer 1

Let u=ln t, e^u=t, du=dt/t, dt=t du
Then:
∫ln² t dt = ∫t u² du = ∫u² e^u du
Integrating by parts:
u² e^u - ∫2u e^u du
u² e^u - 2u e^u + ∫2e^u du
u² e^u - 2u e^u + 2e^u + C
t ln² t - 2t ln t + 2t +C
And we are done.

Answer 2

We know that Integral ln(t) dt = t*ln(t) -t

In yiur integral, put u = ln(t) and dv = ln(t) dt. Then, by parts, Int (ln(t)) dt = uv - Integral vdu = ln(t)*[t*ln(t) -t ] - Integral ((t*ln(t)-t)/t) = ln(t)*[t*ln(t) -t ] - Integral (ln(t) -1) dt = ln(t)*[t*ln(t) -t ] - (t*ln(t) - t) - t +C. = ln(t)*[t*ln(t) -t ] - t*ln(t) + C = t*ln(t)^2 -2t*ln(t) + C.

To show that Integral ln(t) dt = t*ln(t) -t , integrate by parts putting u = ln(t) and dv = dx. Then, you get t*ln(t) - Integral t*(1/t) dt = t*ln(t) - t + c

Now, yiub can do some simplifications, f you want