What is the integral of (-38y+25)/(-y(7y-5)) ?

The integral of (-38y+25)/(-y(7y-5)) is 3/7 ln|7y-5|+5ln|y|+C

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integral of (-38y+25)/(-y(7y-5))

\int \frac{- 38 y + 25}{- y ( 7 y - 5 )} d y

\frac{3}{7} ln ∣ 7 y - 5 ∣ + 5 ln ∣ y ∣ + C

Solution steps

\int \frac{- 38 y + 25}{- y ( 7 y - 5 )} d y

Take the constant out:

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

= - \int \frac{- 38 y + 25}{y ( 7 y - 5 )} d y

Expand

\frac{- 38 y + 25}{y ( 7 y - 5 )} : - \frac{38 y}{y ( 7 y - 5 )} + \frac{25}{y ( 7 y - 5 )}

Apply the Sum Rule:

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

= - (- \int \frac{38 y}{y ( 7 y - 5 )} d y + \int \frac{25}{y ( 7 y - 5 )} d y)

\int \frac{38 y}{y ( 7 y - 5 )} d y = \frac{38}{7} ln ∣ 7 y - 5 ∣

\int \frac{25}{y ( 7 y - 5 )} d y = 25 (- \frac{1}{5} ln ∣ y ∣ + \frac{1}{5} ln ∣ 7 y - 5 ∣)

= - (- \frac{38}{7} ln ∣ 7 y - 5 ∣ + 25 (- \frac{1}{5} ln ∣ y ∣ + \frac{1}{5} ln ∣ 7 y - 5 ∣))

Simplify

- (- \frac{38}{7} ln ∣ 7 y - 5 ∣ + 25 (- \frac{1}{5} ln ∣ y ∣ + \frac{1}{5} ln ∣ 7 y - 5 ∣)) : \frac{3}{7} ln ∣ 7 y - 5 ∣ + 5 ln ∣ y ∣

= \frac{3}{7} ln ∣ 7 y - 5 ∣ + 5 ln ∣ y ∣

Add a constant to the solution

= \frac{3}{7} ln ∣ 7 y - 5 ∣ + 5 ln ∣ y ∣ + C

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\frac{d ^{4}}{d x ^{4}} (18 e^{x^{2}})

x \to \infty lim (2^{x} - 1)

\frac{d}{d x} (4 e^{- x^{4}})

What is the integral of (-38y+25)/(-y(7y-5)) ?
The integral of (-38y+25)/(-y(7y-5)) is 3/7 ln|7y-5|+5ln|y|+C