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

The integral of x/(4x^2+4x+5) is 1/8 (ln|4x^2+4x+5|-arctan(x+1/2))+C

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

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

\frac{1}{8} (ln 4 x^{2} + 4 x + 5 - arctan (x + \frac{1}{2})) + C

Solution steps

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

Complete the square

4 x^{2} + 4 x + 5 : 4 (x + \frac{1}{2})^{2} + 4

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

Apply the Sum Rule:

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

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

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

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

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

Substitute back

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

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

Simplify

\frac{1}{2} (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 4 - \frac{1}{4} arctan (x + \frac{1}{2})) : \frac{1}{8} (ln 4 x^{2} + 4 x + 5 - arctan (x + \frac{1}{2}))

= \frac{1}{8} (ln 4 x^{2} + 4 x + 5 - arctan (x + \frac{1}{2}))

Add a constant to the solution

= \frac{1}{8} (ln 4 x^{2} + 4 x + 5 - arctan (x + \frac{1}{2})) + C

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What is the integral of x/(4x^2+4x+5) ?
The integral of x/(4x^2+4x+5) is 1/8 (ln|4x^2+4x+5|-arctan(x+1/2))+C