What is the integral of 1/(x^2+6x+15) ?

The integral of 1/(x^2+6x+15) is 1/(sqrt(6))arctan((x+3)/(sqrt(6)))+C

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integral of 1/(x^2+6x+15)

\int \frac{1}{x ^{2} + 6 x + 15} d x

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

Solution steps

\int \frac{1}{x ^{2} + 6 x + 15} d x

Complete the square

x^{2} + 6 x + 15 : (x + 3)^{2} + 6

= \int \frac{1}{( x + 3 ) ^{2} + 6} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 6} d u

= \int \frac{1}{u ^{2} + 6} 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

y = \frac{1}{2} x + \frac{1}{4} x sin (2 x)

f (x) = x x, at x = 4

\int \frac{( e ^{x} )}{e ^{x} ( e ^{x} - 3 )} d x

\int \frac{( x ^{3} )}{x ^{2} + 81} d x

x^{2} sin (x)

What is the integral of 1/(x^2+6x+15) ?
The integral of 1/(x^2+6x+15) is 1/(sqrt(6))arctan((x+3)/(sqrt(6)))+C