What is the integral of (x(y^2-5y+6))/(x^2+1) ?

The integral of (x(y^2-5y+6))/(x^2+1) is (6+y^2-5y) 1/2 ln|1+x^2|+C

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integral of (x(y^2-5y+6))/(x^2+1)

\int \frac{x ( y ^{2} - 5 y + 6 )}{x ^{2} + 1} d x

(6 + y^{2} - 5 y) \frac{1}{2} ln 1 + x^{2} + C

Solution steps

\int \frac{x ( y ^{2} - 5 y + 6 )}{x ^{2} + 1} d x

Take the constant out:

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

= (6 + y^{2} - 5 y) \cdot \int \frac{x}{1 + x ^{2}} d x

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

= (6 + y^{2} - 5 y) \frac{1}{2} ln 1 + x^{2}

Add a constant to the solution

= (6 + y^{2} - 5 y) \frac{1}{2} ln 1 + x^{2} + C

What is the integral of (x(y^2-5y+6))/(x^2+1) ?
The integral of (x(y^2-5y+6))/(x^2+1) is (6+y^2-5y) 1/2 ln|1+x^2|+C