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

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

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

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

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

Solution steps

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

Complete the square

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

= \int \frac{1}{2 ( x + \frac{1}{4} ) ^{2} + \frac{7}{8}} d x

Apply u-substitution

: \int \frac{8}{16 u ^{2} + 7} d u

= \int \frac{8}{16 u ^{2} + 7} d u

Take the constant out:

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

= 8 \cdot \int \frac{1}{16 u ^{2} + 7} d u

Apply Integral Substitution

= 8 \cdot \int \frac{1}{4 7 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 8 \cdot \frac{1}{4 7} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 8 \cdot \frac{1}{4 7} arctan (v)

Substitute back

= 8 \cdot \frac{1}{4 7} arctan (\frac{4}{7} (x + \frac{1}{4}))

Simplify

8 \cdot \frac{1}{4 7} arctan (\frac{4}{7} (x + \frac{1}{4})) : \frac{2}{7} arctan (\frac{4 x + 1}{7})

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (t^{6} e^{- x})

y^{^{''}} + 4 y = 4 x

\int x^{2} + 5 x + 1 d x

\frac{d}{d x} f (x) = (x - 1 - x^{2})^{2}

\frac{d}{d x} (12 cos (2 x + \frac{π}{6}))

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