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

The integral of x/(2x^2+x+3) is 2(1/8 ln|16x^2+8x+24|-1/(4sqrt(23))arctan((4x+1)/(sqrt(23))))+C

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

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

2 (\frac{1}{8} ln 16 x^{2} + 8 x + 24 - \frac{1}{4 23} arctan (\frac{4 x + 1}{23})) + C

Solution steps

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

Complete the square

2 x^{2} + x + 3 : 2 (x + \frac{1}{4})^{2} + \frac{23}{8}

= \int \frac{x}{2 ( x + \frac{1}{4} ) ^{2} + \frac{23}{8}} d x

Apply u-substitution

= \int \frac{2 ( 4 u - 1 )}{16 u ^{2} + 23} 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} + 23} d u

Expand

\frac{4 u - 1}{16 u ^{2} + 23} : \frac{4 u}{16 u ^{2} + 23} - \frac{1}{16 u ^{2} + 23}

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} + 23} d u - \int \frac{1}{16 u ^{2} + 23} d u)

\int \frac{4 u}{16 u ^{2} + 23} d u = \frac{1}{8} ln 16 u^{2} + 23

\int \frac{1}{16 u ^{2} + 23} d u = \frac{1}{4 23} arctan (\frac{4}{23} u)

= 2 (\frac{1}{8} ln 16 u^{2} + 23 - \frac{1}{4 23} arctan (\frac{4}{23} u))

Substitute back

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

= 2 (\frac{1}{8} ln 16 (x + \frac{1}{4})^{2} + 23 - \frac{1}{4 23} arctan (\frac{4}{23} (x + \frac{1}{4})))

Simplify

2 (\frac{1}{8} ln 16 (x + \frac{1}{4})^{2} + 23 - \frac{1}{4 23} arctan (\frac{4}{23} (x + \frac{1}{4}))) : 2 (\frac{1}{8} ln 16 x^{2} + 8 x + 24 - \frac{1}{4 23} arctan (\frac{4 x + 1}{23}))

= 2 (\frac{1}{8} ln 16 x^{2} + 8 x + 24 - \frac{1}{4 23} arctan (\frac{4 x + 1}{23}))

Add a constant to the solution

= 2 (\frac{1}{8} ln 16 x^{2} + 8 x + 24 - \frac{1}{4 23} arctan (\frac{4 x + 1}{23})) + C

What is the integral of x/(2x^2+x+3) ?
The integral of x/(2x^2+x+3) is 2(1/8 ln|16x^2+8x+24|-1/(4sqrt(23))arctan((4x+1)/(sqrt(23))))+C