What is the integral of ((x+3))/((x^2+4x+7)) ?

The integral of ((x+3))/((x^2+4x+7)) is 1/2 ln|x^2+4x+7|+1/(sqrt(3))arctan((x+2)/(sqrt(3)))+C

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integral of ((x+3))/((x^2+4x+7))

integral

\frac{( x + 3 )}{( x ^{2} + 4 x + 7 )}

\frac{1}{2} ln x^{2} + 4 x + 7 + \frac{1}{3} arctan (\frac{x + 2}{3}) + C

Solution steps

\int \frac{x + 3}{x ^{2} + 4 x + 7} d x

Apply u-substitution

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

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

Complete the square

u^{2} - 2 u + 4 : (u - 1)^{2} + 3

= \int \frac{u}{( u - 1 ) ^{2} + 3} d u

Apply u-substitution

: \int \frac{v + 1}{v ^{2} + 3} d v

= \int \frac{v + 1}{v ^{2} + 3} d v

Expand

\frac{v + 1}{v ^{2} + 3} : \frac{v}{v ^{2} + 3} + \frac{1}{v ^{2} + 3}

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} + 3} d v + \int \frac{1}{v ^{2} + 3} d v

\int \frac{v}{v ^{2} + 3} d v = \frac{1}{2} ln v^{2} + 3

\int \frac{1}{v ^{2} + 3} d v = \frac{1}{3} arctan (\frac{v}{3})

= \frac{1}{2} ln v^{2} + 3 + \frac{1}{3} arctan (\frac{v}{3})

Substitute back

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

Simplify

\frac{1}{2} ln (x + 3 - 1)^{2} + 3 + \frac{1}{3} arctan (\frac{x + 3 - 1}{3}) : \frac{1}{2} ln x^{2} + 4 x + 7 + \frac{1}{3} arctan (\frac{x + 2}{3})

= \frac{1}{2} ln x^{2} + 4 x + 7 + \frac{1}{3} arctan (\frac{x + 2}{3})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 7 + \frac{1}{3} arctan (\frac{x + 2}{3}) + C

2 x^{2} + 1

e^{- 1} f (x)

tan^{5} (x)

x^{2} (1 + x^{3})

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

What is the integral of ((x+3))/((x^2+4x+7)) ?
The integral of ((x+3))/((x^2+4x+7)) is 1/2 ln|x^2+4x+7|+1/(sqrt(3))arctan((x+2)/(sqrt(3)))+C