What is the integral of ((4x+5))/(x^2+4x+20) ?

The integral of ((4x+5))/(x^2+4x+20) is 1/4 (8ln| 1/16 (x+2)^2+1|-3arctan(1/4 (x+2)))+C

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integral of ((4x+5))/(x^2+4x+20)

\int \frac{( 4 x + 5 )}{x ^{2} + 4 x + 20} d x

\frac{1}{4} (8 ln \frac{1}{16} (x + 2)^{2} + 1 - 3 arctan (\frac{1}{4} (x + 2))) + C

Solution steps

\int \frac{4 x + 5}{x ^{2} + 4 x + 20} d x

Complete the square

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

= \int \frac{4 x + 5}{( x + 2 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{4 u - 3}{u ^{2} + 16} d u

= \int \frac{4 u - 3}{u ^{2} + 16} d u

Apply Integral Substitution

= \int \frac{16 v - 3}{4 ( 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}{4} \cdot \int \frac{16 v - 3}{v ^{2} + 1} d v

Expand

\frac{16 v - 3}{v ^{2} + 1} : \frac{16 v}{v ^{2} + 1} - \frac{3}{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}{4} (\int \frac{16 v}{v ^{2} + 1} d v - \int \frac{3}{v ^{2} + 1} d v)

\int \frac{16 v}{v ^{2} + 1} d v = 8 ln v^{2} + 1

\int \frac{3}{v ^{2} + 1} d v = 3 arctan (v)

= \frac{1}{4} (8 ln v^{2} + 1 - 3 arctan (v))

Substitute back

= \frac{1}{4} (8 ln (\frac{x + 2}{4})^{2} + 1 - 3 arctan (\frac{x + 2}{4}))

Simplify

\frac{1}{4} (8 ln (\frac{x + 2}{4})^{2} + 1 - 3 arctan (\frac{x + 2}{4})) : \frac{1}{4} (8 ln \frac{1}{16} (x + 2)^{2} + 1 - 3 arctan (\frac{1}{4} (x + 2)))

= \frac{1}{4} (8 ln \frac{1}{16} (x + 2)^{2} + 1 - 3 arctan (\frac{1}{4} (x + 2)))

Add a constant to the solution

= \frac{1}{4} (8 ln \frac{1}{16} (x + 2)^{2} + 1 - 3 arctan (\frac{1}{4} (x + 2))) + C

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What is the integral of ((4x+5))/(x^2+4x+20) ?
The integral of ((4x+5))/(x^2+4x+20) is 1/4 (8ln| 1/16 (x+2)^2+1|-3arctan(1/4 (x+2)))+C