What is the integral of (2x+5)/(2x^2+2x+1) ?

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

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integral of (2x+5)/(2x^2+2x+1)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u}{u ^{2} - 8 u + 17} d u

= \int \frac{u}{u ^{2} - 8 u + 17} d u

Complete the square

u^{2} - 8 u + 17 : (u - 4)^{2} + 1

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

Apply u-substitution

: \int \frac{v + 4}{v ^{2} + 1} d v

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

Expand

\frac{v + 4}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} + \frac{4}{v ^{2} + 1}

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{v}{v ^{2} + 1} d v + \int \frac{4}{v ^{2} + 1} d v

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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