What is the integral of ((2x+7))/(2x^2+2x+1) ?

The integral of ((2x+7))/(2x^2+2x+1) is 1/2 ln|4x^2+4x+2|+6arctan(2x+1)+C

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integral of ((2x+7))/(2x^2+2x+1)

\int \frac{( 2 x + 7 )}{2 x ^{2} + 2 x + 1} d x

\frac{1}{2} ln 4 x^{2} + 4 x + 2 + 6 arctan (2 x + 1) + C

Solution steps

\int \frac{2 x + 7}{2 x ^{2} + 2 x + 1} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 12 u + 37} d u

= \int \frac{u}{u ^{2} - 12 u + 37} d u

Complete the square

u^{2} - 12 u + 37 : (u - 6)^{2} + 1

= \int \frac{u}{( u - 6 ) ^{2} + 1} d u

Apply u-substitution

: \int \frac{v + 6}{v ^{2} + 1} d v

= \int \frac{v + 6}{v ^{2} + 1} d v

Expand

\frac{v + 6}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} + \frac{6}{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{6}{v ^{2} + 1} d v

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

\int \frac{6}{v ^{2} + 1} d v = 6 arctan (v)

= \frac{1}{2} ln v^{2} + 1 + 6 arctan (v)

Substitute back

= \frac{1}{2} ln (2 x + 7 - 6)^{2} + 1 + 6 arctan (2 x + 7 - 6)

Simplify

\frac{1}{2} ln (2 x + 7 - 6)^{2} + 1 + 6 arctan (2 x + 7 - 6) : \frac{1}{2} ln 4 x^{2} + 4 x + 2 + 6 arctan (2 x + 1)

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

Add a constant to the solution

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

\int cos (2) x^{2} d x

f (x) = \frac{7}{x}, at x = 6

y^{^{'}} = \frac{e ^{x} - e ^{y}}{1 + e ^{x}}

\int \frac{sin ( 2 x )}{12 + cos ^{2} ( x )} d x

\frac{2}{( s ^{3} - 7 s ^{2} + 2 s + 40 )}

What is the integral of ((2x+7))/(2x^2+2x+1) ?
The integral of ((2x+7))/(2x^2+2x+1) is 1/2 ln|4x^2+4x+2|+6arctan(2x+1)+C