What is the integral of 1/(x^2+x+5) ?

The integral of 1/(x^2+x+5) is 2/(sqrt(19))arctan((2x+1)/(sqrt(19)))+C

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integral of 1/(x^2+x+5)

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

\frac{2}{19} arctan (\frac{2 x + 1}{19}) + C

Solution steps

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

Complete the square

x^{2} + x + 5 : (x + \frac{1}{2})^{2} + \frac{19}{4}

= \int \frac{1}{( x + \frac{1}{2} ) ^{2} + \frac{19}{4}} d x

Apply u-substitution

: \int \frac{4}{4 u ^{2} + 19} d u

= \int \frac{4}{4 u ^{2} + 19} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 u ^{2} + 19} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{2 19 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2 19} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 4 \cdot \frac{1}{2 19} arctan (v)

Substitute back

= 4 \cdot \frac{1}{2 19} arctan (\frac{2}{19} (x + \frac{1}{2}))

Simplify

4 \cdot \frac{1}{2 19} arctan (\frac{2}{19} (x + \frac{1}{2})) : \frac{2}{19} arctan (\frac{2 x + 1}{19})

= \frac{2}{19} arctan (\frac{2 x + 1}{19})

Add a constant to the solution

= \frac{2}{19} arctan (\frac{2 x + 1}{19}) + C

\int e^{3 x - 2} d x

\frac{d y}{d x} - 9 y^{2} e^{- 1 x} = 0, y (0) = 5

y = x^{4}, at x = 0

cos (5 t)

\int x (\frac{1}{4} e^{- \frac{x}{4}}) d x

What is the integral of 1/(x^2+x+5) ?
The integral of 1/(x^2+x+5) is 2/(sqrt(19))arctan((2x+1)/(sqrt(19)))+C