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

x \to - 1 lim (\frac{x - 3}{x + 1})

f (x) = \frac{x ^{3} + 2 x ^{2} + 3 x + 4}{x}

\frac{d}{d x} (9 - x)

4 π x^{2}

n = 1 \sum \infty \frac{1}{n ^{2} + 2 n}

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