What is the integral of [xln(x^2+1)+(x^2+1)y] ?

The integral of [xln(x^2+1)+(x^2+1)y] is y(x+(x^3)/3)+1/2 (ln(1+x^2)+x^2ln(1+x^2)-1-x^2)+C

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integral of [xln(x^2+1)+(x^2+1)y]

\int [x ln (x^{2} + 1) + (x^{2} + 1) y] d x

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

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int (1 + x^{2}) y d x = y (x + \frac{x ^{3}}{3})

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

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

Add a constant to the solution

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

8 9 x^{2} + 5

\frac{\partial}{\partial y} (9 - x^{2} - 3 y)

\frac{\partial}{\partial x} (2 sin (6 x) cos (4 y))

y = \frac{x}{5} - 4

\frac{d}{d x} (\frac{4}{x ^{4}} - 4 5 x)

What is the integral of [xln(x^2+1)+(x^2+1)y] ?
The integral of [xln(x^2+1)+(x^2+1)y] is y(x+(x^3)/3)+1/2 (ln(1+x^2)+x^2ln(1+x^2)-1-x^2)+C