What is the integral of x^3ln(1+x) ?

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

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integral of x^3ln(1+x)

\int x^{3} ln (1 + x) d x

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

Solution steps

\int x^{3} ln (1 + x) d x

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

f (x) = (4 x + 1)^{2}, at x = 0

(x + 1) \cdot ln (x)

\int \frac{1}{e ^{x}} cos (π) x d x

y^{^{'}} = 10 x y

\int \frac{1}{20} (x + 1)^{2} (x - 3) (x + 4) d x

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