What is the integral of x^4ln(1+x) ?

The integral of x^4ln(1+x) is 1/5 x^5ln(1+x)-1/5 ((x^5)/5-(x^4)/4+(x^3)/3-(x^2)/2+x-ln|x+1|)+C

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integral of x^4ln(1+x)

\int x^{4} ln (1 + x) d x

\frac{1}{5} x^{5} ln (1 + x) - \frac{1}{5} (\frac{x ^{5}}{5} - \frac{x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x - ln ∣ x + 1 ∣) + C

Solution steps

\int x^{4} ln (1 + x) d x

Apply Integration By Parts

= \frac{1}{5} x^{5} ln (1 + x) - \int \frac{x ^{5}}{5 ( 1 + x )} d x

\int \frac{x ^{5}}{5 ( 1 + x )} d x = \frac{1}{5} (\frac{x ^{5}}{5} - \frac{x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x - ln ∣ x + 1 ∣)

= \frac{1}{5} x^{5} ln (1 + x) - \frac{1}{5} (\frac{x ^{5}}{5} - \frac{x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x - ln ∣ x + 1 ∣)

Add a constant to the solution

= \frac{1}{5} x^{5} ln (1 + x) - \frac{1}{5} (\frac{x ^{5}}{5} - \frac{x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x - ln ∣ x + 1 ∣) + C

What is the integral of x^4ln(1+x) ?
The integral of x^4ln(1+x) is 1/5 x^5ln(1+x)-1/5 ((x^5)/5-(x^4)/4+(x^3)/3-(x^2)/2+x-ln|x+1|)+C