What is the integral of ((e^2*y))/((4-y)) ?

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

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integral of ((e^2*y))/((4-y))

integral

\frac{( e ^{2} \cdot y )}{( 4 - y )}

- e^{2} (- 4 + y + 4 ln ∣ 4 - y ∣) + C

Solution steps

\int \frac{e ^{2} y}{4 - y} d y

Take the constant out:

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

= e^{2} \cdot \int \frac{y}{4 - y} d y

Apply u-substitution

: \int - \frac{- u + 4}{u} d u

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

Take the constant out:

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

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

Expand

\frac{- u + 4}{u} : - 1 + \frac{4}{u}

= e^{2} (- \int - 1 + \frac{4}{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

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

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= e^{2} (- (- u + 4 ln ∣ u ∣))

Substitute back

u = 4 - y

= e^{2} (- (- (4 - y) + 4 ln ∣ 4 - y ∣))

Simplify

= - e^{2} (- 4 + y + 4 ln ∣ 4 - y ∣)

Add a constant to the solution

= - e^{2} (- 4 + y + 4 ln ∣ 4 - y ∣) + C

x^{3} (e^{- x})^{4}

- t + 3

(arccos (x))^{2}

\frac{1}{( s + 1 ) ^{2} + 1}

\frac{1}{( v + 1 ) ^{2}}

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