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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2} d v

= 2 \cdot \int \frac{sec ( v )}{2} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 2 \cdot \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 2 \cdot \frac{1}{2} ln tan (arctan (\frac{2}{3} (y + \frac{1}{2}))) + sec (arctan (\frac{2}{3} (y + \frac{1}{2})))

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (e^{l n (c o s (x^{2}))} 2 x + 1)

\frac{d y}{d x} + 2 x y = 5 x

(- 4, - 6), (- 2, 0)

n = 1 \sum \infty n^{5} (1 + n^{6})^{6}

\frac{d}{d x} (\frac{f ( x )}{x ^{5}})

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