What is the integral of x/(x^2+4x+20) ?

The integral of x/(x^2+4x+20) is 1/2 (ln| 1/16 (x+2)^2+1|-arctan(1/4 (x+2)))+C

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integral of x/(x^2+4x+20)

\int \frac{x}{x ^{2} + 4 x + 20} d x

\frac{1}{2} (ln \frac{1}{16} (x + 2)^{2} + 1 - arctan (\frac{1}{4} (x + 2))) + C

Solution steps

\int \frac{x}{x ^{2} + 4 x + 20} d x

Complete the square

x^{2} + 4 x + 20 : (x + 2)^{2} + 16

= \int \frac{x}{( x + 2 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{u - 2}{u ^{2} + 16} d u

= \int \frac{u - 2}{u ^{2} + 16} d u

Apply Integral Substitution

= \int \frac{2 v - 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{2 v - 1}{v ^{2} + 1} d v

Expand

\frac{2 v - 1}{v ^{2} + 1} : \frac{2 v}{v ^{2} + 1} - \frac{1}{v ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{2 v}{v ^{2} + 1} d v - \int \frac{1}{v ^{2} + 1} d v)

\int \frac{2 v}{v ^{2} + 1} d v = ln v^{2} + 1

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

= \frac{1}{2} (ln v^{2} + 1 - arctan (v))

Substitute back

= \frac{1}{2} (ln (\frac{x + 2}{4})^{2} + 1 - arctan (\frac{x + 2}{4}))

Simplify

\frac{1}{2} (ln (\frac{x + 2}{4})^{2} + 1 - arctan (\frac{x + 2}{4})) : \frac{1}{2} (ln \frac{1}{16} (x + 2)^{2} + 1 - arctan (\frac{1}{4} (x + 2)))

= \frac{1}{2} (ln \frac{1}{16} (x + 2)^{2} + 1 - arctan (\frac{1}{4} (x + 2)))

Add a constant to the solution

= \frac{1}{2} (ln \frac{1}{16} (x + 2)^{2} + 1 - arctan (\frac{1}{4} (x + 2))) + C

\frac{\partial}{\partial y} (cos (x) cos (y))

\int e^{4 x + 8} d x

f (x) = \frac{2 x + 1}{x - 1}

\frac{d}{d x} (\frac{\frac{d}{d x} ( ln ( x ) )}{x})

What is the integral of x/(x^2+4x+20) ?
The integral of x/(x^2+4x+20) is 1/2 (ln| 1/16 (x+2)^2+1|-arctan(1/4 (x+2)))+C