What is the integral of 1/(x^2+8x+20) ?

The integral of 1/(x^2+8x+20) is 1/2 arctan((x+4)/2)+C

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integral of 1/(x^2+8x+20)

\int \frac{1}{x ^{2} + 8 x + 20} d x

\frac{1}{2} arctan (\frac{x + 4}{2}) + C

Solution steps

\int \frac{1}{x ^{2} + 8 x + 20} d x

Complete the square

x^{2} + 8 x + 20 : (x + 4)^{2} + 4

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

Add a constant to the solution

= \frac{1}{2} arctan (\frac{x + 4}{2}) + C

x \to 0 lim (\frac{4}{x} - \frac{4}{∣ x ∣})

y^{^{'}} + y = 5 sin (2 x)

\frac{\partial}{\partial x} (sin (n x) e^{- n^{2} t})

x \to \infty lim (1 - e^{x})

x \to - 1 lim (x^{2} + 4)

What is the integral of 1/(x^2+8x+20) ?
The integral of 1/(x^2+8x+20) is 1/2 arctan((x+4)/2)+C