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

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

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integral of (y+1)/(y^2+y+1)

\int \frac{y + 1}{y ^{2} + y + 1} d y

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

Solution steps

\int \frac{y + 1}{y ^{2} + y + 1} d y

Complete the square

y^{2} + y + 1 : (y + \frac{1}{2})^{2} + \frac{3}{4}

= \int \frac{y + 1}{( y + \frac{1}{2} ) ^{2} + \frac{3}{4}} d y

Apply u-substitution

: \int \frac{2 ( 2 u + 1 )}{4 u ^{2} + 3} d u

= \int \frac{2 ( 2 u + 1 )}{4 u ^{2} + 3} 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} + 3} d u

Expand

\frac{2 u + 1}{4 u ^{2} + 3} : \frac{2 u}{4 u ^{2} + 3} + \frac{1}{4 u ^{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

= 2 (\int \frac{2 u}{4 u ^{2} + 3} d u + \int \frac{1}{4 u ^{2} + 3} d u)

\int \frac{2 u}{4 u ^{2} + 3} d u = \frac{1}{4} ln 4 u^{2} + 3

\int \frac{1}{4 u ^{2} + 3} d u = \frac{1}{2 3} arctan (\frac{2}{3} u)

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

Substitute back

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

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

Simplify

2 (\frac{1}{4} ln 4 (y + \frac{1}{2})^{2} + 3 + \frac{1}{2 3} arctan (\frac{2}{3} (y + \frac{1}{2}))) : \frac{1}{2} ln 4 y^{2} + 4 y + 4 + \frac{1}{3} arctan (\frac{2 y + 1}{3})

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

Add a constant to the solution

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

\frac{\partial}{\partial z} (x cos (y) sin (z))

\int \frac{x ^{3}}{x - 1} d x

x \to 2 lim (\frac{x + 6}{x - 2})

\int 3 x^{2} (2 x^{3} + 5) d x

\frac{d y}{d x} = - k y

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