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

The integral of (4x^2-3x+2)/(4x^2-4x+3) is x+1/8 (ln|4x^2-4x+3|-sqrt(2)arctan(sqrt(2)(x-1/2)))+C

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

\int \frac{4 x ^{2} - 3 x + 2}{4 x ^{2} - 4 x + 3} d x

x + \frac{1}{8} (ln 4 x^{2} - 4 x + 3 - 2 arctan (2 (x - \frac{1}{2}))) + C

Solution steps

\int \frac{4 x ^{2} - 3 x + 2}{4 x ^{2} - 4 x + 3} d x

Long division

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

= \int 1 + \frac{x - 1}{4 x ^{2} - 4 x + 3} 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 1 d x + \int \frac{x - 1}{4 x ^{2} - 4 x + 3} d x

\int 1 d x = x

\int \frac{x - 1}{4 x ^{2} - 4 x + 3} d x = \frac{1}{8} (ln 4 x^{2} - 4 x + 3 - 2 arctan (2 (x - \frac{1}{2})))

= x + \frac{1}{8} (ln 4 x^{2} - 4 x + 3 - 2 arctan (2 (x - \frac{1}{2})))

Add a constant to the solution

= x + \frac{1}{8} (ln 4 x^{2} - 4 x + 3 - 2 arctan (2 (x - \frac{1}{2}))) + C

\int_{1}^{2} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

2 cos (x) + cos^{2} (x)

\int 4 + x d x

y^{^{''}} + 3 y^{^{'}} + 2 y = 6

x \to 6 lim (\frac{1}{x - 2})

What is the integral of (4x^2-3x+2)/(4x^2-4x+3) ?
The integral of (4x^2-3x+2)/(4x^2-4x+3) is x+1/8 (ln|4x^2-4x+3|-sqrt(2)arctan(sqrt(2)(x-1/2)))+C