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

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

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integral of 3/(9x^2+6x+1)

\int \frac{3}{9 x ^{2} + 6 x + 1} d x

- \frac{1}{3 x + 1} + C

Solution steps

\int \frac{3}{9 x ^{2} + 6 x + 1} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{9 x ^{2} + 6 x + 1} d x

Complete the square

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

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

Apply u-substitution

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

= 3 \cdot \int \frac{1}{9 u ^{2}} d u

Take the constant out:

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

= 3 \cdot \frac{1}{9} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 3 \cdot \frac{1}{9} (- \frac{1}{u})

Substitute back

u = x + \frac{1}{3}

= 3 \cdot \frac{1}{9} (- \frac{1}{x + \frac{1}{3}})

Simplify

3 \cdot \frac{1}{9} (- \frac{1}{x + \frac{1}{3}}) : - \frac{1}{3 x + 1}

= - \frac{1}{3 x + 1}

Add a constant to the solution

= - \frac{1}{3 x + 1} + C

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

\frac{3000 - 10 x ^{2} + 1}{3 x ^{3}}

\frac{1}{2} sin (2 t)

\frac{d}{d x} (\frac{1 + x}{1 - x})

7 x - 3

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