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

The integral of x/(x^2+x+4) is 1/2 ln|4x^2+4x+16|-1/(sqrt(15))arctan((2x+1)/(sqrt(15)))+C

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

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

\frac{1}{2} ln 4 x^{2} + 4 x + 16 - \frac{1}{15} arctan (\frac{2 x + 1}{15}) + C

Solution steps

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

Complete the square

x^{2} + x + 4 : (x + \frac{1}{2})^{2} + \frac{15}{4}

= \int \frac{x}{( x + \frac{1}{2} ) ^{2} + \frac{15}{4}} d x

Apply u-substitution

= \int \frac{2 ( 2 u - 1 )}{4 u ^{2} + 15} d u

Take the constant out:

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

= 2 \cdot \int \frac{2 u - 1}{4 u ^{2} + 15} d u

Expand

\frac{2 u - 1}{4 u ^{2} + 15} : \frac{2 u}{4 u ^{2} + 15} - \frac{1}{4 u ^{2} + 15}

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{2 u}{4 u ^{2} + 15} d u - \int \frac{1}{4 u ^{2} + 15} d u)

\int \frac{2 u}{4 u ^{2} + 15} d u = \frac{1}{4} ln 4 u^{2} + 15

\int \frac{1}{4 u ^{2} + 15} d u = \frac{1}{2 15} arctan (\frac{2}{15} u)

= 2 (\frac{1}{4} ln 4 u^{2} + 15 - \frac{1}{2 15} arctan (\frac{2}{15} u))

Substitute back

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

= 2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 15 - \frac{1}{2 15} arctan (\frac{2}{15} (x + \frac{1}{2})))

Simplify

2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 15 - \frac{1}{2 15} arctan (\frac{2}{15} (x + \frac{1}{2}))) : \frac{1}{2} ln 4 x^{2} + 4 x + 16 - \frac{1}{15} arctan (\frac{2 x + 1}{15})

= \frac{1}{2} ln 4 x^{2} + 4 x + 16 - \frac{1}{15} arctan (\frac{2 x + 1}{15})

Add a constant to the solution

= \frac{1}{2} ln 4 x^{2} + 4 x + 16 - \frac{1}{15} arctan (\frac{2 x + 1}{15}) + C

What is the integral of x/(x^2+x+4) ?
The integral of x/(x^2+x+4) is 1/2 ln|4x^2+4x+16|-1/(sqrt(15))arctan((2x+1)/(sqrt(15)))+C