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

The integral of t^2*ln^2(t) is 1/3 t^3ln^2(t)-2/3 (1/3 t^3ln(t)-(t^3)/9)+C

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integral of t^2*ln^2(t)

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

\frac{1}{3} t^{3} ln^{2} (t) - \frac{2}{3} (\frac{1}{3} t^{3} ln (t) - \frac{t ^{3}}{9}) + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{3} t^{3} ln^{2} (t) - \int \frac{2}{3} t^{2} ln (t) d t

= \frac{1}{3} t^{3} ln^{2} (t) - \int \frac{2}{3} t^{2} ln (t) d t

\int \frac{2}{3} t^{2} ln (t) d t = \frac{2}{3} (\frac{1}{3} t^{3} ln (t) - \frac{t ^{3}}{9})

= \frac{1}{3} t^{3} ln^{2} (t) - \frac{2}{3} (\frac{1}{3} t^{3} ln (t) - \frac{t ^{3}}{9})

Add a constant to the solution

= \frac{1}{3} t^{3} ln^{2} (t) - \frac{2}{3} (\frac{1}{3} t^{3} ln (t) - \frac{t ^{3}}{9}) + C

x \to 0 lim (x^{- 3})

\frac{d}{d x} (4 x^{- \frac{1}{3}} + \frac{9}{x ^{\frac{5}{3}}})

\frac{d y}{d x}, x cos (2 x + 3 y) = y sin (x)

\int 2 t - t - \frac{9}{t ^{2}} d t

x \to 0 lim (x^{2} - 4)

What is the integral of t^2*ln^2(t) ?
The integral of t^2*ln^2(t) is 1/3 t^3ln^2(t)-2/3 (1/3 t^3ln(t)-(t^3)/9)+C