What is the integral of ((y-3))/(y^2-6y+1) ?

The integral of ((y-3))/(y^2-6y+1) is 1/2 ln|y^2-6y+1|+C

Popular >

integral of ((y-3))/(y^2-6y+1)

\int \frac{( y - 3 )}{y ^{2} - 6 y + 1} d y

\frac{1}{2} ln y^{2} - 6 y + 1 + C

Solution steps

\int \frac{y - 3}{y ^{2} - 6 y + 1} d y

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = y^{2} - 6 y + 1

= \frac{1}{2} ln y^{2} - 6 y + 1

Add a constant to the solution

= \frac{1}{2} ln y^{2} - 6 y + 1 + C

\int cos^{6} (\frac{x}{2}) sin^{4} (\frac{x}{2}) d x

\int \frac{sin ( θ )}{cos ^{2} ( θ ) + cos ( θ ) - 2} d θ

\int 32 arctan (x) d x

\int x^{2} 3 x^{3} - 2 d x

\int (x^{3} - 7 x) \cdot e^{5 x} d x

What is the integral of ((y-3))/(y^2-6y+1) ?
The integral of ((y-3))/(y^2-6y+1) is 1/2 ln|y^2-6y+1|+C