What is the integral of ((y-4))/(y^2-8y+10) ?

The integral of ((y-4))/(y^2-8y+10) is 1/2 ln|y^2-8y+10|+C

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integral of ((y-4))/(y^2-8y+10)

\int \frac{( y - 4 )}{y ^{2} - 8 y + 10} d y

\frac{1}{2} ln y^{2} - 8 y + 10 + C

Solution steps

\int \frac{y - 4}{y ^{2} - 8 y + 10} d y

Apply u-substitution

: \int \frac{u}{u ^{2} - 6} d u

= \int \frac{u}{u ^{2} - 6} d u

Apply u-substitution

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

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

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}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

= \frac{1}{2} ln (y - 4)^{2} - 6

Simplify

\frac{1}{2} ln (y - 4)^{2} - 6 : \frac{1}{2} ln y^{2} - 8 y + 10

= \frac{1}{2} ln y^{2} - 8 y + 10

Add a constant to the solution

= \frac{1}{2} ln y^{2} - 8 y + 10 + C

f (x) = \frac{5}{x}, at x = a

f (x) = x^{4} + 4 x^{2} - x, at x = 1

\int sin^{5} (7 x) cos^{2} (7 x) d x

\int_{- 1}^{1} \frac{5 x}{( 4 + x ^{2} ) ^{2}} d x

What is the integral of ((y-4))/(y^2-8y+10) ?
The integral of ((y-4))/(y^2-8y+10) is 1/2 ln|y^2-8y+10|+C