What is the integral of 1/(x^2+6x+109) ?

The integral of 1/(x^2+6x+109) is 1/10 arctan((x+3)/(10))+C

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integral of 1/(x^2+6x+109)

\int \frac{1}{x ^{2} + 6 x + 109} d x

\frac{1}{10} arctan (\frac{x + 3}{10}) + C

Solution steps

\int \frac{1}{x ^{2} + 6 x + 109} d x

Complete the square

x^{2} + 6 x + 109 : (x + 3)^{2} + 100

= \int \frac{1}{( x + 3 ) ^{2} + 100} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 100} d u

= \int \frac{1}{u ^{2} + 100} d u

Apply Integral Substitution

= \int \frac{1}{10 ( 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}{10} \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}{10} arctan (v)

Substitute back

= \frac{1}{10} arctan (\frac{x + 3}{10})

Add a constant to the solution

= \frac{1}{10} arctan (\frac{x + 3}{10}) + C

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What is the integral of 1/(x^2+6x+109) ?
The integral of 1/(x^2+6x+109) is 1/10 arctan((x+3)/(10))+C