What is the integral of 1/(x^2+4x+40) ?

The integral of 1/(x^2+4x+40) is 1/6 arctan((x+2)/6)+C

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integral of 1/(x^2+4x+40)

\int \frac{1}{x ^{2} + 4 x + 40} d x

\frac{1}{6} arctan (\frac{x + 2}{6}) + C

Solution steps

\int \frac{1}{x ^{2} + 4 x + 40} d x

Complete the square

x^{2} + 4 x + 40 : (x + 2)^{2} + 36

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

Add a constant to the solution

= \frac{1}{6} arctan (\frac{x + 2}{6}) + C

\int \frac{1}{x ^{2} - 9} d x

P (t) = 25 e^{0.3 t}

\frac{\partial ^{2}}{\partial x \partial y} (x arcsin (\frac{y}{x}))

\frac{2}{s ^{2} + 2 s + 2}

What is the integral of 1/(x^2+4x+40) ?
The integral of 1/(x^2+4x+40) is 1/6 arctan((x+2)/6)+C