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

The integral of 1/(y^2-3y+2) is -ln|2y-2|+ln|2y-4|+C

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

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

- ln ∣ 2 y - 2 ∣ + ln ∣ 2 y - 4 ∣ + C

Solution steps

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

Complete the square

y^{2} - 3 y + 2 : (y - \frac{3}{2})^{2} - \frac{1}{4}

= \int \frac{1}{( y - \frac{3}{2} ) ^{2} - \frac{1}{4}} d y

Apply u-substitution

: \int \frac{4}{4 u ^{2} - 1} d u

= \int \frac{4}{4 u ^{2} - 1} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 u ^{2} - 1} d u

Factor

4 u^{2} - 1 : - (- 4 u^{2} + 1)

= 4 \cdot \int \frac{1}{- ( - 4 u ^{2} + 1 )} d u

Take the constant out:

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

= 4 (- \int \frac{1}{- 4 u ^{2} + 1} d u)

Apply Integral Substitution

= 4 (- \int \frac{1}{2 ( - v ^{2} + 1 )} d v)

Take the constant out:

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

= 4 (- \frac{1}{2} \cdot \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 4 (- \frac{1}{2} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 4 (- \frac{1}{2} (\frac{ln 2 ( y - \frac{3}{2} ) + 1}{2} - \frac{ln 2 ( y - \frac{3}{2} ) - 1}{2}))

Simplify

4 (- \frac{1}{2} (\frac{ln 2 ( y - \frac{3}{2} ) + 1}{2} - \frac{ln 2 ( y - \frac{3}{2} ) - 1}{2})) : - ln ∣ 2 y - 2 ∣ + ln ∣ 2 y - 4 ∣

= - ln ∣ 2 y - 2 ∣ + ln ∣ 2 y - 4 ∣

Add a constant to the solution

= - ln ∣ 2 y - 2 ∣ + ln ∣ 2 y - 4 ∣ + C

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

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

\int 0.03 e^{- 0.03 x} d x

\frac{\partial}{\partial y} (arctan (yz))

e^{- x} (y^{^{'}} - y) = y^{2}

What is the integral of 1/(y^2-3y+2) ?
The integral of 1/(y^2-3y+2) is -ln|2y-2|+ln|2y-4|+C