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

Popular >

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

x \to 5 lim (in (x^{2} - 25))

\int \frac{1}{x + 1 - x} d x

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

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

\frac{d}{d x} (arccsc (- 5 x^{2}))

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