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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\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}))) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{2 u - 1}{2 ( 4 u ^{2} + 2 )} d u

= \int \frac{2 u - 1}{2 ( 4 u ^{2} + 2 )} d u

Take the constant out:

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

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

Expand

\frac{2 u - 1}{4 u ^{2} + 2} : \frac{2 u}{4 u ^{2} + 2} - \frac{1}{4 u ^{2} + 2}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{2 u}{4 u ^{2} + 2} d u - \int \frac{1}{4 u ^{2} + 2} d u)

\int \frac{2 u}{4 u ^{2} + 2} d u = \frac{1}{4} ln 4 u^{2} + 2

\int \frac{1}{4 u ^{2} + 2} d u = \frac{1}{2 2} arctan (2 u)

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

Substitute back

u = x - \frac{1}{2}

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} = \frac{2 x + 1}{x ^{2} - 3 x + 2}, y (3) = 0

x \frac{d y}{d x} = 1 + 3 y + y^{2}

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

- x^{2} + 7 x + 2, at x = 3

y = 2 x^{3} - x^{2} + 1, (1, 2)

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