What is the integral of ((2x+3))/(4x^2+4x+5) ?

The integral of ((2x+3))/(4x^2+4x+5) is 1/4 ln|4x^2+4x+5|+1/2 arctan(x+1/2)+C

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integral of ((2x+3))/(4x^2+4x+5)

\int \frac{( 2 x + 3 )}{4 x ^{2} + 4 x + 5} d x

\frac{1}{4} ln 4 x^{2} + 4 x + 5 + \frac{1}{2} arctan (x + \frac{1}{2}) + C

Solution steps

\int \frac{2 x + 3}{4 x ^{2} + 4 x + 5} d x

Expand

\frac{2 x + 3}{4 x ^{2} + 4 x + 5} : \frac{2 x}{4 x ^{2} + 4 x + 5} + \frac{3}{4 x ^{2} + 4 x + 5}

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{2 x}{4 x ^{2} + 4 x + 5} d x + \int \frac{3}{4 x ^{2} + 4 x + 5} d x

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

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

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

Add similar elements:

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

= \frac{1}{4} ln 4 x^{2} + 4 x + 5 + \frac{1}{2} arctan (x + \frac{1}{2})

Add a constant to the solution

= \frac{1}{4} ln 4 x^{2} + 4 x + 5 + \frac{1}{2} arctan (x + \frac{1}{2}) + C

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What is the integral of ((2x+3))/(4x^2+4x+5) ?
The integral of ((2x+3))/(4x^2+4x+5) is 1/4 ln|4x^2+4x+5|+1/2 arctan(x+1/2)+C