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

The integral of (1+y^2)/(2-y) is -1/2 (2-y)^2+8-4y-5ln|2-y|+C

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

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

- \frac{1}{2} (2 - y)^{2} + 8 - 4 y - 5 ln ∣ 2 - y ∣ + C

Solution steps

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

Apply u-substitution

: \int - \frac{u ^{2} - 4 u + 5}{u} d u

= \int - \frac{u ^{2} - 4 u + 5}{u} d u

Take the constant out:

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

= - \int \frac{u ^{2} - 4 u + 5}{u} d u

Expand

\frac{u ^{2} - 4 u + 5}{u} : u - 4 + \frac{5}{u}

= - \int u - 4 + \frac{5}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int u d u - \int 4 d u + \int \frac{5}{u} d u)

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

\int 4 d u = 4 u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= - (\frac{u ^{2}}{2} - 4 u + 5 ln ∣ u ∣)

Substitute back

u = 2 - y

= - (\frac{( 2 - y ) ^{2}}{2} - 4 (2 - y) + 5 ln ∣ 2 - y ∣)

Simplify

- (\frac{( 2 - y ) ^{2}}{2} - 4 (2 - y) + 5 ln ∣ 2 - y ∣) : - \frac{1}{2} (2 - y)^{2} + 8 - 4 y - 5 ln ∣ 2 - y ∣

= - \frac{1}{2} (2 - y)^{2} + 8 - 4 y - 5 ln ∣ 2 - y ∣

Add a constant to the solution

= - \frac{1}{2} (2 - y)^{2} + 8 - 4 y - 5 ln ∣ 2 - y ∣ + C

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What is the integral of (1+y^2)/(2-y) ?
The integral of (1+y^2)/(2-y) is -1/2 (2-y)^2+8-4y-5ln|2-y|+C