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

The integral of t(ln(t))^2 is 1/2 t^2ln^2(t)-1/2 t^2ln(t)+(t^2)/4+C

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integral of t(ln(t))^2

\int t (ln (t))^{2} d t

\frac{1}{2} t^{2} ln^{2} (t) - \frac{1}{2} t^{2} ln (t) + \frac{t ^{2}}{4} + C

Solution steps

\int t (ln (t))^{2} d t

Apply Integration By Parts

: \frac{1}{2} t^{2} ln^{2} (t) - \int t ln (t) d t

= \frac{1}{2} t^{2} ln^{2} (t) - \int t ln (t) d t

\int t ln (t) d t = \frac{1}{2} t^{2} ln (t) - \frac{t ^{2}}{4}

= \frac{1}{2} t^{2} ln^{2} (t) - (\frac{1}{2} t^{2} ln (t) - \frac{t ^{2}}{4})

Simplify

= \frac{1}{2} t^{2} ln^{2} (t) - \frac{1}{2} t^{2} ln (t) + \frac{t ^{2}}{4}

Add a constant to the solution

= \frac{1}{2} t^{2} ln^{2} (t) - \frac{1}{2} t^{2} ln (t) + \frac{t ^{2}}{4} + C

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What is the integral of t(ln(t))^2 ?
The integral of t(ln(t))^2 is 1/2 t^2ln^2(t)-1/2 t^2ln(t)+(t^2)/4+C