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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

y = 2^{3 x^{2}}

\int \frac{( cos ^{5} ( t ) )}{sin ( t )} d t

x \to a lim (\frac{f ( x )}{g ( x )})

\frac{d}{d x} (2 cos (\frac{x}{2}) + π)

n = 0 \sum \infty (- \frac{5}{9})^{n}

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