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

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

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integral of x^2ln^2(x)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

y^{^{'}} + 3 x^{2} y = x e^{- x^{3}}

y^{^{'}} = \frac{3 sec ( y )}{( x + 5 ) ^{2}}

f (x) = 2^{30}

\frac{d}{d x} (lo g_{e} (\frac{4 x ^{2} - 9}{2 x - 3}))

\int_{- \infty}^{\infty} x^{23} e^{- x^{24}} d x

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