What is the integral of (y^2+3)/(y-4) ?

The integral of (y^2+3)/(y-4) is ((y-4)^2)/2+8(y-4)+19ln|y-4|+C

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integral of (y^2+3)/(y-4)

\int \frac{y ^{2} + 3}{y - 4} d y

\frac{( y - 4 ) ^{2}}{2} + 8 (y - 4) + 19 ln ∣ y - 4 ∣ + C

Solution steps

\int \frac{y ^{2} + 3}{y - 4} d y

Apply u-substitution

= \int \frac{u ^{2} + 8 u + 19}{u} d u

Expand

\frac{u ^{2} + 8 u + 19}{u} : u + 8 + \frac{19}{u}

= \int u + 8 + \frac{19}{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 8 d u + \int \frac{19}{u} d u

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

\int 8 d u = 8 u

\int \frac{19}{u} d u = 19 ln ∣ u ∣

= \frac{u ^{2}}{2} + 8 u + 19 ln ∣ u ∣

Substitute back

u = y - 4

= \frac{( y - 4 ) ^{2}}{2} + 8 (y - 4) + 19 ln ∣ y - 4 ∣

Add a constant to the solution

= \frac{( y - 4 ) ^{2}}{2} + 8 (y - 4) + 19 ln ∣ y - 4 ∣ + C

What is the integral of (y^2+3)/(y-4) ?
The integral of (y^2+3)/(y-4) is ((y-4)^2)/2+8(y-4)+19ln|y-4|+C