What is the integral of (1-3x+6x^2)/((x^2+1)(4x+1)) ?

The integral of (1-3x+6x^2)/((x^2+1)(4x+1)) is 1/2 ln|x^2+1|-arctan(x)+1/2 ln|4x+1|+C

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integral of (1-3x+6x^2)/((x^2+1)(4x+1))

\int \frac{1 - 3 x + 6 x ^{2}}{( x ^{2} + 1 ) ( 4 x + 1 )} d x

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

Solution steps

\int \frac{1 - 3 x + 6 x ^{2}}{( x ^{2} + 1 ) ( 4 x + 1 )} d x

Take the partial fraction of

\frac{1 - 3 x + 6 x ^{2}}{( x ^{2} + 1 ) ( 4 x + 1 )} : \frac{x - 1}{x ^{2} + 1} + \frac{2}{4 x + 1}

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

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

\int \frac{2}{4 x + 1} d x = \frac{1}{2} ln ∣ 4 x + 1 ∣

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

Add a constant to the solution

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

x \to 0 lim (x cot (\frac{π x}{2}))

\int (5 + x^{8})^{4} 8 x^{7} d x

\int x^{3} sin (3 x) d x

\frac{d y}{d x} + y = e^{- x}

\frac{5 s}{s ^{2} ( s ^{2} + 16 )}

What is the integral of (1-3x+6x^2)/((x^2+1)(4x+1)) ?
The integral of (1-3x+6x^2)/((x^2+1)(4x+1)) is 1/2 ln|x^2+1|-arctan(x)+1/2 ln|4x+1|+C