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

The integral of x/(x^2+x+2) is 1/2 ln|4x^2+4x+8|-1/(sqrt(7))arctan((2x+1)/(sqrt(7)))+C

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

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

\frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{1}{7} arctan (\frac{2 x + 1}{7}) + C

Solution steps

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

Complete the square

x^{2} + x + 2 : (x + \frac{1}{2})^{2} + \frac{7}{4}

= \int \frac{x}{( x + \frac{1}{2} ) ^{2} + \frac{7}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u - 1 )}{4 u ^{2} + 7} d u

= \int \frac{2 ( 2 u - 1 )}{4 u ^{2} + 7} d u

Take the constant out:

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

= 2 \cdot \int \frac{2 u - 1}{4 u ^{2} + 7} d u

Expand

\frac{2 u - 1}{4 u ^{2} + 7} : \frac{2 u}{4 u ^{2} + 7} - \frac{1}{4 u ^{2} + 7}

Apply the Sum Rule:

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

= 2 (\int \frac{2 u}{4 u ^{2} + 7} d u - \int \frac{1}{4 u ^{2} + 7} d u)

\int \frac{2 u}{4 u ^{2} + 7} d u = \frac{1}{4} ln 4 u^{2} + 7

\int \frac{1}{4 u ^{2} + 7} d u = \frac{1}{2 7} arctan (\frac{2}{7} u)

= 2 (\frac{1}{4} ln 4 u^{2} + 7 - \frac{1}{2 7} arctan (\frac{2}{7} u))

Substitute back

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

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

Simplify

2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 7 - \frac{1}{2 7} arctan (\frac{2}{7} (x + \frac{1}{2}))) : \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{1}{7} arctan (\frac{2 x + 1}{7})

= \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{1}{7} arctan (\frac{2 x + 1}{7})

Add a constant to the solution

= \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{1}{7} arctan (\frac{2 x + 1}{7}) + C

\frac{d}{d x} (\frac{2 x ^{3} + 5}{4 x ^{2} + 7})

\frac{d}{d x} ((f (x) (x^{3}))^{2})

\frac{d}{d x} (3 cot (2 x) sec (2 x))

\frac{d ^{2}}{d x ^{2}} (x^{4} - 2 x^{3})

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

What is the integral of x/(x^2+x+2) ?
The integral of x/(x^2+x+2) is 1/2 ln|4x^2+4x+8|-1/(sqrt(7))arctan((2x+1)/(sqrt(7)))+C