What is the integral of (12x^2-6x+2)/((4x+1)(x^2+1)) ?

The integral of (12x^2-6x+2)/((4x+1)(x^2+1)) is ln|4x+1|+ln|x^2+1|-2arctan(x)+C

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integral of (12x^2-6x+2)/((4x+1)(x^2+1))

\int \frac{12 x ^{2} - 6 x + 2}{( 4 x + 1 ) ( x ^{2} + 1 )} d x

ln ∣ 4 x + 1 ∣ + ln x^{2} + 1 - 2 arctan (x) + C

Solution steps

\int \frac{12 x ^{2} - 6 x + 2}{( 4 x + 1 ) ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{12 x ^{2} - 6 x + 2}{( 4 x + 1 ) ( x ^{2} + 1 )} : \frac{4}{4 x + 1} + \frac{2 x - 2}{x ^{2} + 1}

= \int \frac{4}{4 x + 1} + \frac{2 x - 2}{x ^{2} + 1} d x

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

\int \frac{4}{4 x + 1} d x = ln ∣ 4 x + 1 ∣

\int \frac{2 x - 2}{x ^{2} + 1} d x = ln x^{2} + 1 - 2 arctan (x)

= ln ∣ 4 x + 1 ∣ + ln x^{2} + 1 - 2 arctan (x)

Add a constant to the solution

= ln ∣ 4 x + 1 ∣ + ln x^{2} + 1 - 2 arctan (x) + C

What is the integral of (12x^2-6x+2)/((4x+1)(x^2+1)) ?
The integral of (12x^2-6x+2)/((4x+1)(x^2+1)) is ln|4x+1|+ln|x^2+1|-2arctan(x)+C