What is the integral of (6x-1)/(4x^2+4x+10) ?

The integral of (6x-1)/(4x^2+4x+10) is 3/4 ln|4x^2+4x+10|-2/3 arctan(2/3 x+1/3)+C

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integral of (6x-1)/(4x^2+4x+10)

\int \frac{6 x - 1}{4 x ^{2} + 4 x + 10} d x

\frac{3}{4} ln 4 x^{2} + 4 x + 10 - \frac{2}{3} arctan (\frac{2}{3} x + \frac{1}{3}) + C

Solution steps

\int \frac{6 x - 1}{4 x ^{2} + 4 x + 10} d x

Expand

\frac{6 x - 1}{4 x ^{2} + 4 x + 10} : \frac{6 x}{4 x ^{2} + 4 x + 10} - \frac{1}{4 x ^{2} + 4 x + 10}

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{6 x}{4 x ^{2} + 4 x + 10} d x - \int \frac{1}{4 x ^{2} + 4 x + 10} d x

\int \frac{6 x}{4 x ^{2} + 4 x + 10} d x = 3 (\frac{1}{4} ln 4 x^{2} + 4 x + 10 - \frac{1}{6} arctan (\frac{2}{3} x + \frac{1}{3}))

\int \frac{1}{4 x ^{2} + 4 x + 10} d x = \frac{1}{6} arctan (\frac{2}{3} x + \frac{1}{3})

= 3 (\frac{1}{4} ln 4 x^{2} + 4 x + 10 - \frac{1}{6} arctan (\frac{2}{3} x + \frac{1}{3})) - \frac{1}{6} arctan (\frac{2}{3} x + \frac{1}{3})

Simplify

3 (\frac{1}{4} ln 4 x^{2} + 4 x + 10 - \frac{1}{6} arctan (\frac{2}{3} x + \frac{1}{3})) - \frac{1}{6} arctan (\frac{2}{3} x + \frac{1}{3}) : \frac{3}{4} ln 4 x^{2} + 4 x + 10 - \frac{2}{3} arctan (\frac{2}{3} x + \frac{1}{3})

= \frac{3}{4} ln 4 x^{2} + 4 x + 10 - \frac{2}{3} arctan (\frac{2}{3} x + \frac{1}{3})

Add a constant to the solution

= \frac{3}{4} ln 4 x^{2} + 4 x + 10 - \frac{2}{3} arctan (\frac{2}{3} x + \frac{1}{3}) + C

\int \frac{28}{( 196 x ^{2} + 1 ) ^{2}} d x

y^{'} = y (5 - y)

\frac{\partial}{\partial y} (- 4 e^{x} cos (yz))

\frac{d}{d x} (1 + x^{2} + x^{2} 1 + x^{2})

\int (5 e^{2 x} + \frac{1}{x}) d x

What is the integral of (6x-1)/(4x^2+4x+10) ?
The integral of (6x-1)/(4x^2+4x+10) is 3/4 ln|4x^2+4x+10|-2/3 arctan(2/3 x+1/3)+C