What is the integral of y/(y+3) ?

The integral of y/(y+3) is y+3-3ln|y+3|+C

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integral of y/(y+3)

\int \frac{y}{y + 3} d y

y + 3 - 3 ln ∣ y + 3 ∣ + C

Solution steps

\int \frac{y}{y + 3} d y

Apply u-substitution

: \int \frac{u - 3}{u} d u

= \int \frac{u - 3}{u} d u

Expand

\frac{u - 3}{u} : 1 - \frac{3}{u}

= \int 1 - \frac{3}{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 1 d u - \int \frac{3}{u} d u

\int 1 d u = u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

= u - 3 ln ∣ u ∣

Substitute back

u = y + 3

= y + 3 - 3 ln ∣ y + 3 ∣

Add a constant to the solution

= y + 3 - 3 ln ∣ y + 3 ∣ + C

x \to 0 lim (\frac{5 x ^{2}}{x ^{4}})

\int_{0}^{2} x^{3} 3 x + 1 d x

ln (x), at 4

f (x) = \frac{- x ( x + 4 )}{( x ^{2} - x - 2 ) ^{2}}

n = 1 \sum \infty \frac{( e ^{n} )}{n}