What is the integral of 1/((y^2+6y+5)) ?

The integral of 1/((y^2+6y+5)) is 1/4 ln|y+1|-1/4 ln|y+5|+C

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integral of 1/((y^2+6y+5))

integral

\frac{1}{( y ^{2} + 6 y + 5 )}

\frac{1}{4} ln ∣ y + 1 ∣ - \frac{1}{4} ln ∣ y + 5 ∣ + C

Solution steps

\int \frac{1}{y ^{2} + 6 y + 5} d y

Take the partial fraction of

\frac{1}{y ^{2} + 6 y + 5} : \frac{1}{4 ( y + 1 )} - \frac{1}{4 ( y + 5 )}

= \int \frac{1}{4 ( y + 1 )} - \frac{1}{4 ( y + 5 )} d y

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}{4 ( y + 1 )} d y - \int \frac{1}{4 ( y + 5 )} d y

\int \frac{1}{4 ( y + 1 )} d y = \frac{1}{4} ln ∣ y + 1 ∣

\int \frac{1}{4 ( y + 5 )} d y = \frac{1}{4} ln ∣ y + 5 ∣

= \frac{1}{4} ln ∣ y + 1 ∣ - \frac{1}{4} ln ∣ y + 5 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ y + 1 ∣ - \frac{1}{4} ln ∣ y + 5 ∣ + C

What is the integral of 1/((y^2+6y+5)) ?
The integral of 1/((y^2+6y+5)) is 1/4 ln|y+1|-1/4 ln|y+5|+C