What is the integral of 1/(x^2+6x+13) ?

The integral of 1/(x^2+6x+13) is 1/2 arctan((x+3)/2)+C

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integral of 1/(x^2+6x+13)

\int \frac{1}{x ^{2} + 6 x + 13} d x

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

Solution steps

\int \frac{1}{x ^{2} + 6 x + 13} d x

Complete the square

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

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

Apply u-substitution

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

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

Apply Integral Substitution

= \int \frac{1}{2 ( 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}{2} \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}{2} arctan (v)

Substitute back

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

Add a constant to the solution

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

\int (\frac{1}{x}) d x

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

\frac{d}{d x} (e^{- x} \cdot sin (x))

\frac{d}{d x} (e^{\frac{1}{x ^{3}}})

\frac{d y}{d x} y = (y + 1) (2 x + 1), y (- 1) = 1

What is the integral of 1/(x^2+6x+13) ?
The integral of 1/(x^2+6x+13) is 1/2 arctan((x+3)/2)+C