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

The integral of y/((y+5)^2(y-5)) is -1/20 ln|y+5|-1/(2(y+5))+1/20 ln|y-5|+C

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

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

- \frac{1}{20} ln ∣ y + 5 ∣ - \frac{1}{2 ( y + 5 )} + \frac{1}{20} ln ∣ y - 5 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 5}{u ^{2} ( u - 10 )} d u

= \int \frac{u - 5}{u ^{2} ( u - 10 )} d u

Take the partial fraction of

\frac{u - 5}{u ^{2} ( u - 10 )} : - \frac{1}{20 u} + \frac{1}{2 u ^{2}} + \frac{1}{20 ( u - 10 )}

= \int - \frac{1}{20 u} + \frac{1}{2 u ^{2}} + \frac{1}{20 ( u - 10 )} 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 \frac{1}{20 u} d u + \int \frac{1}{2 u ^{2}} d u + \int \frac{1}{20 ( u - 10 )} d u

\int \frac{1}{20 u} d u = \frac{1}{20} ln ∣ u ∣

\int \frac{1}{2 u ^{2}} d u = - \frac{1}{2 u}

\int \frac{1}{20 ( u - 10 )} d u = \frac{1}{20} ln ∣ u - 10 ∣

= - \frac{1}{20} ln ∣ u ∣ - \frac{1}{2 u} + \frac{1}{20} ln ∣ u - 10 ∣

Substitute back

u = y + 5

= - \frac{1}{20} ln ∣ y + 5 ∣ - \frac{1}{2 ( y + 5 )} + \frac{1}{20} ln ∣ y + 5 - 10 ∣

Simplify

- \frac{1}{20} ln ∣ y + 5 ∣ - \frac{1}{2 ( y + 5 )} + \frac{1}{20} ln ∣ y + 5 - 10 ∣ : - \frac{1}{20} ln ∣ y + 5 ∣ - \frac{1}{2 ( y + 5 )} + \frac{1}{20} ln ∣ y - 5 ∣

= - \frac{1}{20} ln ∣ y + 5 ∣ - \frac{1}{2 ( y + 5 )} + \frac{1}{20} ln ∣ y - 5 ∣

Add a constant to the solution

= - \frac{1}{20} ln ∣ y + 5 ∣ - \frac{1}{2 ( y + 5 )} + \frac{1}{20} ln ∣ y - 5 ∣ + C

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What is the integral of y/((y+5)^2(y-5)) ?
The integral of y/((y+5)^2(y-5)) is -1/20 ln|y+5|-1/(2(y+5))+1/20 ln|y-5|+C