What is the integral of y/((y+7)(2y-1)) ?

The integral of y/((y+7)(2y-1)) is 7/15 ln|y+7|+1/30 ln|2y-1|+C

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integral of y/((y+7)(2y-1))

\int \frac{y}{( y + 7 ) ( 2 y - 1 )} d y

\frac{7}{15} ln ∣ y + 7 ∣ + \frac{1}{30} ln ∣ 2 y - 1 ∣ + C

Solution steps

\int \frac{y}{( y + 7 ) ( 2 y - 1 )} d y

Take the partial fraction of

\frac{y}{( y + 7 ) ( 2 y - 1 )} : \frac{7}{15 ( y + 7 )} + \frac{1}{15 ( 2 y - 1 )}

= \int \frac{7}{15 ( y + 7 )} + \frac{1}{15 ( 2 y - 1 )} 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{7}{15 ( y + 7 )} d y + \int \frac{1}{15 ( 2 y - 1 )} d y

\int \frac{7}{15 ( y + 7 )} d y = \frac{7}{15} ln ∣ y + 7 ∣

\int \frac{1}{15 ( 2 y - 1 )} d y = \frac{1}{30} ln ∣ 2 y - 1 ∣

= \frac{7}{15} ln ∣ y + 7 ∣ + \frac{1}{30} ln ∣ 2 y - 1 ∣

Add a constant to the solution

= \frac{7}{15} ln ∣ y + 7 ∣ + \frac{1}{30} ln ∣ 2 y - 1 ∣ + C

- y^{^{''}} = 1

(t^{2} + 16) y^{^{'}} + 2 t y = t^{2} (t^{2} + 16)

\int \frac{5 x ^{3} - 9 x}{x} d x

\int \frac{25}{( x - 1 ) ( x ^{2} + 4 )} d x

\int (x^{2} + 1)^{2} x d x

What is the integral of y/((y+7)(2y-1)) ?
The integral of y/((y+7)(2y-1)) is 7/15 ln|y+7|+1/30 ln|2y-1|+C