What is the integral of t ln(1+t) dt?

Answer 1

This integral takes a couple steps to get the answer. First, since its a product of an easily integrable and an easily differentiable function, integration by parts makes sense. The integrable function is set to dv:

dv = t dt so: v = (t^2)/2
u = ln(1+t) so: du = dt/(1+t)

Integrating by parts: S u dv = uv - S v du

S t ln(1+t) dy = t^2 ln(1+t)/2 - S (t^2)/(2(1+t)) dt

At least the log is gone but now you need to integrate:

(t^2)/(1 + t)

This can be reduced by trying to get factors of t+1 in the numerator. The numerator could be rewritten as:

t^2 = (t + 1)^2 - 2t - 1

So:

(t^2)/(t+1) = (t+1) - (2t + 1)/((t + 1)

Take the remaining numerator and do a similar trick:

(2t + 1)/(t + 1) = 2(t + 1)/(t + 1) - 1/(t + 1) = 2 - 1/(t+1)

So the integrand reduces to:


(t^2)/(t+1) = (t+1) - 2 + 1/(t+1) = t -1 + 1/(t+1)

Now that's easy to integrate:

S(t^2)/(t+1) dt = S( t -1 + 1/(t+1)) dt = (t^2)/2 - t + ln(t+1)

Combine that with the integration by parts from above:

S t ln(1+t) dy = t^2 ln(1+t)/2 - S (t^2)/(2(1+t)) dt
S t ln(1+t) dy = t^2 ln(1+t)/2 - ((t^2)/2 - t + ln(t+1))/2
S t ln(1+t) dy = (t^2 - 1) ln(1 + t)/2 - (t^2)/4 + t/2

You can prove this is correct by differentiatiing.

The previous answer by oregfiu is correct as well except for a typo. The last 2 characters are "/x" and they should be "/4"

Answer 2

Write ∫t*ln(1 + t)dt = ∫ln(1 + t)*tdt and integrate by parts:

Then ∫ln(1 + t)*tdt = ln(1 + t)*(½*t²) - ∫(½*t²)*1/(1 + t)dt + C

= ½t²*ln(1 + t) - ½∫t²/(1 + t)dt + C

= ½t²*ln(1 + t) - ½∫[t - 1 + 1/(1 + t)]dt + C

= ½t²*ln(1 + t) - ½[½t² - t + ln(1 + t)] + C

= ½t²*ln(1 + t) - t²/4 + ½t - ½ln(1 + t) + C.

Answer 3

= ( 2 (x^2 - 1) * ln(x+1) - (x-2) * x) / x
-