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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x + 3

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

Add a constant to the solution

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

x^{3} (x^{2} + 1)^{7}

x^{2} sin^{9} (+ x cos^{- 2} (x))

\int 3 x^{2} + 2 x - 3 d x

(4 y + t^{2} y) d y - (2 t + t y^{2}) d t = 0

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