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

The integral of x/(2x^2-x+1) is 2(1/8 ln|16x^2-8x+8|+1/(4sqrt(7))arctan((4x-1)/(sqrt(7))))+C

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

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

2 (\frac{1}{8} ln 16 x^{2} - 8 x + 8 + \frac{1}{4 7} arctan (\frac{4 x - 1}{7})) + C

Solution steps

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

Complete the square

2 x^{2} - x + 1 : 2 (x - \frac{1}{4})^{2} + \frac{7}{8}

= \int \frac{x}{2 ( x - \frac{1}{4} ) ^{2} + \frac{7}{8}} d x

Apply u-substitution

: \int \frac{2 ( 4 u + 1 )}{16 u ^{2} + 7} d u

= \int \frac{2 ( 4 u + 1 )}{16 u ^{2} + 7} d u

Take the constant out:

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

= 2 \cdot \int \frac{4 u + 1}{16 u ^{2} + 7} d u

Expand

\frac{4 u + 1}{16 u ^{2} + 7} : \frac{4 u}{16 u ^{2} + 7} + \frac{1}{16 u ^{2} + 7}

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{4 u}{16 u ^{2} + 7} d u + \int \frac{1}{16 u ^{2} + 7} d u)

\int \frac{4 u}{16 u ^{2} + 7} d u = \frac{1}{8} ln 16 u^{2} + 7

\int \frac{1}{16 u ^{2} + 7} d u = \frac{1}{4 7} arctan (\frac{4}{7} u)

= 2 (\frac{1}{8} ln 16 u^{2} + 7 + \frac{1}{4 7} arctan (\frac{4}{7} u))

Substitute back

u = x - \frac{1}{4}

= 2 (\frac{1}{8} ln 16 (x - \frac{1}{4})^{2} + 7 + \frac{1}{4 7} arctan (\frac{4}{7} (x - \frac{1}{4})))

Simplify

2 (\frac{1}{8} ln 16 (x - \frac{1}{4})^{2} + 7 + \frac{1}{4 7} arctan (\frac{4}{7} (x - \frac{1}{4}))) : 2 (\frac{1}{8} ln 16 x^{2} - 8 x + 8 + \frac{1}{4 7} arctan (\frac{4 x - 1}{7}))

= 2 (\frac{1}{8} ln 16 x^{2} - 8 x + 8 + \frac{1}{4 7} arctan (\frac{4 x - 1}{7}))

Add a constant to the solution

= 2 (\frac{1}{8} ln 16 x^{2} - 8 x + 8 + \frac{1}{4 7} arctan (\frac{4 x - 1}{7})) + C

\int_{- 1}^{1} \frac{1}{x + 1} d x

\frac{d y}{d t} - 2 t y = 15 t^{2} e^{t^{2}}

\int x^{3} + x d x

\int \frac{2 x}{x ^{4} + 36} d x

\frac{1}{s} \cdot (e^{- 2 \cdot s \cdot π} - 1)

What is the integral of x/(2x^2-x+1) ?
The integral of x/(2x^2-x+1) is 2(1/8 ln|16x^2-8x+8|+1/(4sqrt(7))arctan((4x-1)/(sqrt(7))))+C