What is the integral of ((3x-2))/(x^2+2x+17) ?

The integral of ((3x-2))/(x^2+2x+17) is 1/4 (6ln| 1/16 (x+1)^2+1|-5arctan(1/4 (x+1)))+C

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integral of ((3x-2))/(x^2+2x+17)

\int \frac{( 3 x - 2 )}{x ^{2} + 2 x + 17} d x

\frac{1}{4} (6 ln \frac{1}{16} (x + 1)^{2} + 1 - 5 arctan (\frac{1}{4} (x + 1))) + C

Solution steps

\int \frac{3 x - 2}{x ^{2} + 2 x + 17} d x

Complete the square

x^{2} + 2 x + 17 : (x + 1)^{2} + 16

= \int \frac{3 x - 2}{( x + 1 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{3 u - 5}{u ^{2} + 16} d u

= \int \frac{3 u - 5}{u ^{2} + 16} d u

Apply Integral Substitution

= \int \frac{12 v - 5}{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{12 v - 5}{v ^{2} + 1} d v

Expand

\frac{12 v - 5}{v ^{2} + 1} : \frac{12 v}{v ^{2} + 1} - \frac{5}{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{12 v}{v ^{2} + 1} d v - \int \frac{5}{v ^{2} + 1} d v)

\int \frac{12 v}{v ^{2} + 1} d v = 6 ln v^{2} + 1

\int \frac{5}{v ^{2} + 1} d v = 5 arctan (v)

= \frac{1}{4} (6 ln v^{2} + 1 - 5 arctan (v))

Substitute back

= \frac{1}{4} (6 ln (\frac{x + 1}{4})^{2} + 1 - 5 arctan (\frac{x + 1}{4}))

Simplify

\frac{1}{4} (6 ln (\frac{x + 1}{4})^{2} + 1 - 5 arctan (\frac{x + 1}{4})) : \frac{1}{4} (6 ln \frac{1}{16} (x + 1)^{2} + 1 - 5 arctan (\frac{1}{4} (x + 1)))

= \frac{1}{4} (6 ln \frac{1}{16} (x + 1)^{2} + 1 - 5 arctan (\frac{1}{4} (x + 1)))

Add a constant to the solution

= \frac{1}{4} (6 ln \frac{1}{16} (x + 1)^{2} + 1 - 5 arctan (\frac{1}{4} (x + 1))) + C

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What is the integral of ((3x-2))/(x^2+2x+17) ?
The integral of ((3x-2))/(x^2+2x+17) is 1/4 (6ln| 1/16 (x+1)^2+1|-5arctan(1/4 (x+1)))+C