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

The integral of (2x^3)/(2x^2-4x+3) is (x^2)/2+2x+5/4 ln|2x^2-4x+3|-1/(sqrt(2))arctan(sqrt(2)(x-1))+C

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

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

\frac{x ^{2}}{2} + 2 x + \frac{5}{4} ln 2 x^{2} - 4 x + 3 - \frac{1}{2} arctan (2 (x - 1)) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Long division

\frac{x ^{3}}{2 x ^{2} - 4 x + 3} : \frac{x}{2} + 1 + \frac{5 x - 6}{2 ( 2 x ^{2} - 4 x + 3 )}

= 2 \cdot \int \frac{x}{2} + 1 + \frac{5 x - 6}{2 ( 2 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

= 2 (\int \frac{x}{2} d x + \int 1 d x + \int \frac{5 x - 6}{2 ( 2 x ^{2} - 4 x + 3 )} d x)

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

\int 1 d x = x

\int \frac{5 x - 6}{2 ( 2 x ^{2} - 4 x + 3 )} d x = \frac{1}{2} (\frac{5}{4} ln 2 x^{2} - 4 x + 3 - \frac{1}{2} arctan (2 (x - 1)))

= 2 (\frac{x ^{2}}{4} + x + \frac{1}{2} (\frac{5}{4} ln 2 x^{2} - 4 x + 3 - \frac{1}{2} arctan (2 (x - 1))))

Simplify

2 (\frac{x ^{2}}{4} + x + \frac{1}{2} (\frac{5}{4} ln 2 x^{2} - 4 x + 3 - \frac{1}{2} arctan (2 (x - 1)))) : \frac{x ^{2}}{2} + 2 x + \frac{5}{4} ln 2 x^{2} - 4 x + 3 - \frac{1}{2} arctan (2 (x - 1))

= \frac{x ^{2}}{2} + 2 x + \frac{5}{4} ln 2 x^{2} - 4 x + 3 - \frac{1}{2} arctan (2 (x - 1))

Add a constant to the solution

= \frac{x ^{2}}{2} + 2 x + \frac{5}{4} ln 2 x^{2} - 4 x + 3 - \frac{1}{2} arctan (2 (x - 1)) + C

\int x^{3} d x

\int 6 x e^{x^{2}} d x

P (t) = - 5 t^{2} + 75 t - 30

\frac{\partial}{\partial x} (x^{3} - x y + y^{3})

x \to 1 lim (x^{\frac{7}{1 - x}})

What is the integral of (2x^3)/(2x^2-4x+3) ?
The integral of (2x^3)/(2x^2-4x+3) is (x^2)/2+2x+5/4 ln|2x^2-4x+3|-1/(sqrt(2))arctan(sqrt(2)(x-1))+C