What is the integral of 1/(9x^2+6x+5) ?

The integral of 1/(9x^2+6x+5) is 1/6 arctan(3/2 x+1/2)+C

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integral of 1/(9x^2+6x+5)

\int \frac{1}{9 x ^{2} + 6 x + 5} d x

\frac{1}{6} arctan (\frac{3}{2} x + \frac{1}{2}) + C

Solution steps

\int \frac{1}{9 x ^{2} + 6 x + 5} d x

Complete the square

9 x^{2} + 6 x + 5 : 9 (x + \frac{1}{3})^{2} + 4

= \int \frac{1}{9 ( x + \frac{1}{3} ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{9 u ^{2} + 4} d u

= \int \frac{1}{9 u ^{2} + 4} d u

Apply Integral Substitution

= \int \frac{1}{6 ( 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}{6} \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}{6} arctan (v)

Substitute back

= \frac{1}{6} arctan (\frac{3}{2} (x + \frac{1}{3}))

Expand

\frac{3}{2} (x + \frac{1}{3}) : \frac{3}{2} x + \frac{1}{2}

= \frac{1}{6} arctan (\frac{3}{2} x + \frac{1}{2})

Add a constant to the solution

= \frac{1}{6} arctan (\frac{3}{2} x + \frac{1}{2}) + C

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What is the integral of 1/(9x^2+6x+5) ?
The integral of 1/(9x^2+6x+5) is 1/6 arctan(3/2 x+1/2)+C