What is the integral of x/(x^2-7x+13) ?

The integral of x/(x^2-7x+13) is 1/2 ln|4x^2-28x+52|+7/(sqrt(3))arctan((2x-7)/(sqrt(3)))+C

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integral of x/(x^2-7x+13)

\int \frac{x}{x ^{2} - 7 x + 13} d x

\frac{1}{2} ln 4 x^{2} - 28 x + 52 + \frac{7}{3} arctan (\frac{2 x - 7}{3}) + C

Solution steps

\int \frac{x}{x ^{2} - 7 x + 13} d x

Complete the square

x^{2} - 7 x + 13 : (x - \frac{7}{2})^{2} + \frac{3}{4}

= \int \frac{x}{( x - \frac{7}{2} ) ^{2} + \frac{3}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u + 7 )}{4 u ^{2} + 3} d u

= \int \frac{2 ( 2 u + 7 )}{4 u ^{2} + 3} d u

Take the constant out:

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

= 2 \cdot \int \frac{2 u + 7}{4 u ^{2} + 3} d u

Expand

\frac{2 u + 7}{4 u ^{2} + 3} : \frac{2 u}{4 u ^{2} + 3} + \frac{7}{4 u ^{2} + 3}

Apply the Sum Rule:

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

= 2 (\int \frac{2 u}{4 u ^{2} + 3} d u + \int \frac{7}{4 u ^{2} + 3} d u)

\int \frac{2 u}{4 u ^{2} + 3} d u = \frac{1}{4} ln 4 u^{2} + 3

\int \frac{7}{4 u ^{2} + 3} d u = \frac{7}{2 3} arctan (\frac{2}{3} u)

= 2 (\frac{1}{4} ln 4 u^{2} + 3 + \frac{7}{2 3} arctan (\frac{2}{3} u))

Substitute back

u = x - \frac{7}{2}

= 2 (\frac{1}{4} ln 4 (x - \frac{7}{2})^{2} + 3 + \frac{7}{2 3} arctan (\frac{2}{3} (x - \frac{7}{2})))

Simplify

2 (\frac{1}{4} ln 4 (x - \frac{7}{2})^{2} + 3 + \frac{7}{2 3} arctan (\frac{2}{3} (x - \frac{7}{2}))) : \frac{1}{2} ln 4 x^{2} - 28 x + 52 + \frac{7}{3} arctan (\frac{2 x - 7}{3})

= \frac{1}{2} ln 4 x^{2} - 28 x + 52 + \frac{7}{3} arctan (\frac{2 x - 7}{3})

Add a constant to the solution

= \frac{1}{2} ln 4 x^{2} - 28 x + 52 + \frac{7}{3} arctan (\frac{2 x - 7}{3}) + C

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What is the integral of x/(x^2-7x+13) ?
The integral of x/(x^2-7x+13) is 1/2 ln|4x^2-28x+52|+7/(sqrt(3))arctan((2x-7)/(sqrt(3)))+C