What is the integral of (e^5y+5)/(e^2y-5) ?

The integral of (e^5y+5)/(e^2y-5) is 1/(e^2)(e^3(e^2y-5)+5e^3ln|e^2y-5|+5ln|e^2y-5|)+C

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integral of (e^5y+5)/(e^2y-5)

integral

\frac{e ^{5} y + 5}{e ^{2} y - 5}

\frac{1}{e ^{2}} (e^{3} (e^{2} y - 5) + 5 e^{3} ln e^{2} y - 5 + 5 ln e^{2} y - 5) + C

Solution steps

\int \frac{e ^{5} y + 5}{e ^{2} y - 5} d y

Apply u-substitution

: \int \frac{e ^{3} ( u + 5 ) + 5}{e ^{2} u} d u

= \int \frac{e ^{3} ( u + 5 ) + 5}{e ^{2} u} d u

Take the constant out:

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

= \frac{1}{e ^{2}} \cdot \int \frac{e ^{3} ( u + 5 ) + 5}{u} d u

Expand

\frac{e ^{3} ( u + 5 ) + 5}{u} : e^{3} + \frac{5 e ^{3}}{u} + \frac{5}{u}

= \frac{1}{e ^{2}} \cdot \int e^{3} + \frac{5 e ^{3}}{u} + \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

= \frac{1}{e ^{2}} (\int e^{3} d u + \int \frac{5 e ^{3}}{u} d u + \int \frac{5}{u} d u)

\int e^{3} d u = e^{3} u

\int \frac{5 e ^{3}}{u} d u = 5 e^{3} ln ∣ u ∣

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= \frac{1}{e ^{2}} (e^{3} u + 5 e^{3} ln ∣ u ∣ + 5 ln ∣ u ∣)

Substitute back

u = e^{2} y - 5

= \frac{1}{e ^{2}} (e^{3} (e^{2} y - 5) + 5 e^{3} ln e^{2} y - 5 + 5 ln e^{2} y - 5)

Add a constant to the solution

= \frac{1}{e ^{2}} (e^{3} (e^{2} y - 5) + 5 e^{3} ln e^{2} y - 5 + 5 ln e^{2} y - 5) + C

\frac{1}{x - x ^{2}}

\frac{x}{6}

\frac{arcsin ( x )}{x ^{2}}

\frac{2 y ^{2}}{x ^{3}}

x^{5} (1 - x^{2})^{0.5}

What is the integral of (e^5y+5)/(e^2y-5) ?
The integral of (e^5y+5)/(e^2y-5) is 1/(e^2)(e^3(e^2y-5)+5e^3ln|e^2y-5|+5ln|e^2y-5|)+C