What is the integral of 1/(41y^2+37y) ?

The integral of 1/(41y^2+37y) is 1/37 ln|y|-1/37 ln|41y+37|+C

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integral of 1/(41y^2+37y)

\int \frac{1}{41 y ^{2} + 37 y} d y

\frac{1}{37} ln ∣ y ∣ - \frac{1}{37} ln ∣ 41 y + 37 ∣ + C

Solution steps

\int \frac{1}{41 y ^{2} + 37 y} d y

Take the partial fraction of

\frac{1}{41 y ^{2} + 37 y} : \frac{1}{37 y} - \frac{41}{37 ( 41 y + 37 )}

= \int \frac{1}{37 y} - \frac{41}{37 ( 41 y + 37 )} 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}{37 y} d y - \int \frac{41}{37 ( 41 y + 37 )} d y

\int \frac{1}{37 y} d y = \frac{1}{37} ln ∣ y ∣

\int \frac{41}{37 ( 41 y + 37 )} d y = \frac{1}{37} ln ∣ 41 y + 37 ∣

= \frac{1}{37} ln ∣ y ∣ - \frac{1}{37} ln ∣ 41 y + 37 ∣

Add a constant to the solution

= \frac{1}{37} ln ∣ y ∣ - \frac{1}{37} ln ∣ 41 y + 37 ∣ + C

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What is the integral of 1/(41y^2+37y) ?
The integral of 1/(41y^2+37y) is 1/37 ln|y|-1/37 ln|41y+37|+C