What is the integral of 1/(x^2+6x+45) ?

The integral of 1/(x^2+6x+45) is 1/6 arctan((x+3)/6)+C

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integral of 1/(x^2+6x+45)

\int \frac{1}{x ^{2} + 6 x + 45} d x

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

Solution steps

\int \frac{1}{x ^{2} + 6 x + 45} d x

Complete the square

x^{2} + 6 x + 45 : (x + 3)^{2} + 36

= \int \frac{1}{( x + 3 ) ^{2} + 36} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 36} d u

= \int \frac{1}{u ^{2} + 36} 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{x + 3}{6})

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{( 5 + x ) ^{2}})

\int \frac{1}{( t ^{2} - 1 ) ^{2}} d t

n = 1 \sum \infty \frac{4}{4 n ^{2} - 1}

\frac{\partial}{\partial x} (x^{2} y z^{2} - 4 cos (3 yz))

f (x) = - 3 x^{2} + 1 x, at x = 0

What is the integral of 1/(x^2+6x+45) ?
The integral of 1/(x^2+6x+45) is 1/6 arctan((x+3)/6)+C