What is the integral of (6y-7)/((2y+1)(4y^2+1)) ?

The integral of (6y-7)/((2y+1)(4y^2+1)) is -5/2 ln|2y+1|+5/4 ln|4y^2+1|-arctan(2y)+C

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integral of (6y-7)/((2y+1)(4y^2+1))

\int \frac{6 y - 7}{( 2 y + 1 ) ( 4 y ^{2} + 1 )} d y

- \frac{5}{2} ln ∣ 2 y + 1 ∣ + \frac{5}{4} ln 4 y^{2} + 1 - arctan (2 y) + C

Solution steps

\int \frac{6 y - 7}{( 2 y + 1 ) ( 4 y ^{2} + 1 )} d y

Take the partial fraction of

\frac{6 y - 7}{( 2 y + 1 ) ( 4 y ^{2} + 1 )} : - \frac{5}{2 y + 1} + \frac{10 y - 2}{4 y ^{2} + 1}

= \int - \frac{5}{2 y + 1} + \frac{10 y - 2}{4 y ^{2} + 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{5}{2 y + 1} d y + \int \frac{10 y - 2}{4 y ^{2} + 1} d y

\int \frac{5}{2 y + 1} d y = \frac{5}{2} ln ∣ 2 y + 1 ∣

\int \frac{10 y - 2}{4 y ^{2} + 1} d y = \frac{5}{4} ln 4 y^{2} + 1 - arctan (2 y)

= - \frac{5}{2} ln ∣ 2 y + 1 ∣ + \frac{5}{4} ln 4 y^{2} + 1 - arctan (2 y)

Add a constant to the solution

= - \frac{5}{2} ln ∣ 2 y + 1 ∣ + \frac{5}{4} ln 4 y^{2} + 1 - arctan (2 y) + C

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What is the integral of (6y-7)/((2y+1)(4y^2+1)) ?
The integral of (6y-7)/((2y+1)(4y^2+1)) is -5/2 ln|2y+1|+5/4 ln|4y^2+1|-arctan(2y)+C