What is the integral of x^3ln(x^2+1) ?

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

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integral of x^3ln(x^2+1)

\int x^{3} ln (x^{2} + 1) d x

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

Solution steps

\int x^{3} ln (x^{2} + 1) d x

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

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\frac{\partial}{\partial x} (tan (2 x - y))

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