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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

= 2 (\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)

= 2 (\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))

Simplify

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

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

Add a constant to the solution

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

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