What is the integral of x^3ln^2(x) ?

The integral of x^3ln^2(x) is 1/4 x^4ln^2(x)-1/2 (1/4 x^4ln(x)-(x^4)/(16))+C

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integral of x^3ln^2(x)

\int x^{3} ln^{2} (x) d x

\frac{1}{4} x^{4} ln^{2} (x) - \frac{1}{2} (\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16}) + C

Solution steps

\int x^{3} ln^{2} (x) d x

Apply Integration By Parts

: \frac{1}{4} x^{4} ln^{2} (x) - \int \frac{1}{2} x^{3} ln (x) d x

= \frac{1}{4} x^{4} ln^{2} (x) - \int \frac{1}{2} x^{3} ln (x) d x

\int \frac{1}{2} x^{3} ln (x) d x = \frac{1}{2} (\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16})

= \frac{1}{4} x^{4} ln^{2} (x) - \frac{1}{2} (\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16})

Add a constant to the solution

= \frac{1}{4} x^{4} ln^{2} (x) - \frac{1}{2} (\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16}) + C

\frac{\partial}{\partial x} (y sin (\frac{y}{x}))

x \to - 4 lim (5 x + 2)

x \to - 2 lim (\frac{( 3 )}{( x + 2 )})

\int_{0}^{6} \frac{π}{4} (36 - y^{2}) d y

\int_{0}^{1} 34.6 e^{- 0.06 t} d t

What is the integral of x^3ln^2(x) ?
The integral of x^3ln^2(x) is 1/4 x^4ln^2(x)-1/2 (1/4 x^4ln(x)-(x^4)/(16))+C