What is the integral of 1/(v^2+6v+18) ?

The integral of 1/(v^2+6v+18) is 1/3 arctan((v+3)/3)+C

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integral of 1/(v^2+6v+18)

\int \frac{1}{v ^{2} + 6 v + 18} d v

\frac{1}{3} arctan (\frac{v + 3}{3}) + C

Solution steps

\int \frac{1}{v ^{2} + 6 v + 18} d v

Complete the square

v^{2} + 6 v + 18 : (v + 3)^{2} + 9

= \int \frac{1}{( v + 3 ) ^{2} + 9} d v

Apply u-substitution

: \int \frac{1}{u ^{2} + 9} d u

= \int \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

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

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= \frac{1}{3} arctan (w)

Substitute back

= \frac{1}{3} arctan (\frac{v + 3}{3})

Add a constant to the solution

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

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What is the integral of 1/(v^2+6v+18) ?
The integral of 1/(v^2+6v+18) is 1/3 arctan((v+3)/3)+C