What is the integral of 1/(x^2+2x+6) ?

The integral of 1/(x^2+2x+6) is 1/(sqrt(5))arctan((x+1)/(sqrt(5)))+C

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integral of 1/(x^2+2x+6)

\int \frac{1}{x ^{2} + 2 x + 6} d x

\frac{1}{5} arctan (\frac{x + 1}{5}) + C

Solution steps

\int \frac{1}{x ^{2} + 2 x + 6} d x

Complete the square

x^{2} + 2 x + 6 : (x + 1)^{2} + 5

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 5} d u

= \int \frac{1}{u ^{2} + 5} d u

Apply Integral Substitution

= \int \frac{1}{5 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{5} arctan (v)

Substitute back

= \frac{1}{5} arctan (\frac{x + 1}{5})

Add a constant to the solution

= \frac{1}{5} arctan (\frac{x + 1}{5}) + C

\int (e^{2 t} + \frac{1}{( t + 1 ) ^{2}}) d t

\frac{s - 3}{s ^{2} + 10 s + 9}

x \to 4 lim (2 x + 1)

\int \frac{1}{- u ^{2} + 1} d u

\frac{( x ^{8} )}{8 ( 1 - x ^{2} ) ^{4}}

What is the integral of 1/(x^2+2x+6) ?
The integral of 1/(x^2+2x+6) is 1/(sqrt(5))arctan((x+1)/(sqrt(5)))+C