What is the integral of (2x)/(x^2-4x+13) ?

The integral of (2x)/(x^2-4x+13) is 1/3 (3ln|((x-2)^2)/9+1|+4arctan((x-2)/3))+C

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integral of (2x)/(x^2-4x+13)

\int \frac{2 x}{x ^{2} - 4 x + 13} d x

\frac{1}{3} (3 ln \frac{( x - 2 ) ^{2}}{9} + 1 + 4 arctan (\frac{x - 2}{3})) + C

Solution steps

\int \frac{2 x}{x ^{2} - 4 x + 13} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{x}{x ^{2} - 4 x + 13} d x

Complete the square

x^{2} - 4 x + 13 : (x - 2)^{2} + 9

= 2 \cdot \int \frac{x}{( x - 2 ) ^{2} + 9} d x

Apply u-substitution

= 2 \cdot \int \frac{u + 2}{u ^{2} + 9} d u

Apply Integral Substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{3} \cdot \int \frac{3 v + 2}{v ^{2} + 1} d v

Expand

\frac{3 v + 2}{v ^{2} + 1} : \frac{3 v}{v ^{2} + 1} + \frac{2}{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

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

\int \frac{3 v}{v ^{2} + 1} d v = \frac{3}{2} ln v^{2} + 1

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

= 2 \cdot \frac{1}{3} (\frac{3}{2} ln v^{2} + 1 + 2 arctan (v))

Substitute back

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

Simplify

2 \cdot \frac{1}{3} (\frac{3}{2} ln (\frac{x - 2}{3})^{2} + 1 + 2 arctan (\frac{x - 2}{3})) : \frac{1}{3} (3 ln \frac{( x - 2 ) ^{2}}{9} + 1 + 4 arctan (\frac{x - 2}{3}))

= \frac{1}{3} (3 ln \frac{( x - 2 ) ^{2}}{9} + 1 + 4 arctan (\frac{x - 2}{3}))

Add a constant to the solution

= \frac{1}{3} (3 ln \frac{( x - 2 ) ^{2}}{9} + 1 + 4 arctan (\frac{x - 2}{3})) + C

What is the integral of (2x)/(x^2-4x+13) ?
The integral of (2x)/(x^2-4x+13) is 1/3 (3ln|((x-2)^2)/9+1|+4arctan((x-2)/3))+C