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

The integral of (x+3)/(x^2-x+4) is 1/2 ln|4x^2-4x+16|+7/(sqrt(15))arctan(1/(sqrt(15))(2x-1))+C

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

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

\frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{7}{15} arctan (\frac{1}{15} (2 x - 1)) + C

Solution steps

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

Expand

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

Apply the Sum Rule:

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

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

\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})

\int \frac{3}{x ^{2} - x + 4} d x = \frac{2 3}{5} 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}) + \frac{2 3}{5} arctan (\frac{2 x - 1}{15})

Simplify

\frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{2 x - 1}{15}) + \frac{2 3}{5} arctan (\frac{2 x - 1}{15}) : \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{7}{15} arctan (\frac{1}{15} (2 x - 1))

= \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{7}{15} arctan (\frac{1}{15} (2 x - 1))

Add a constant to the solution

= \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{7}{15} arctan (\frac{1}{15} (2 x - 1)) + C

y^{^{''}} + 2 y^{^{'}} + y = 4 e - t

1 y^{2} - 7 y^{4} (y + 9 y^{3})

x \to 36 lim (\frac{x ^{2}}{9} - \frac{4}{x})

\int \frac{7}{4 + 7 x} d x

\frac{\partial}{\partial x} (3 x^{2} y + x sin (2 y))

What is the integral of (x+3)/(x^2-x+4) ?
The integral of (x+3)/(x^2-x+4) is 1/2 ln|4x^2-4x+16|+7/(sqrt(15))arctan(1/(sqrt(15))(2x-1))+C