What is the integral of xln^4(x) ?

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

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integral of xln^4(x)

\int x ln^{4} (x) d x

\frac{1}{2} x^{2} ln^{4} (x) - x^{2} ln^{3} (x) + \frac{3}{2} x^{2} ln^{2} (x) - \frac{3}{2} x^{2} ln (x) + \frac{3}{4} x^{2} + C

Solution steps

\int x ln^{4} (x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

= \frac{1}{2} x^{2} ln^{4} (x) - x^{2} ln^{3} (x) + \frac{3}{2} x^{2} ln^{2} (x) - \frac{3}{2} x^{2} ln (x) + \frac{3}{4} x^{2} + C

\frac{\partial}{\partial x} (\frac{t}{c ^{2} t ^{2} - x ^{2}})

f (x) = \frac{12}{x + 7}

\frac{\partial}{\partial x} (\frac{2 x - y}{5 x y + 6})

f (t) = arcsin (t)

(1 + x)^{- \frac{1}{2}}

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