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

The integral of x^3(ln(x))^2 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^3(ln(x))^2

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

Simplify

= \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{s}{( s ^{2} + 4 )}

x \to 1 lim ((x^{2} - 2)^{3})

f (x) = 6 sec (x) tan (x)

\int_{0}^{7} \frac{1}{49 + x ^{2}} d x

\int \frac{1}{4 - y ^{2}} d y

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