What is the integral of (3x+1)/(x^2-4x+7) ?

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

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integral of (3x+1)/(x^2-4x+7)

\int \frac{3 x + 1}{x ^{2} - 4 x + 7} d x

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

Solution steps

\int \frac{3 x + 1}{x ^{2} - 4 x + 7} d x

Complete the square

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

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

Apply u-substitution

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

Expand

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

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

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

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

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

Substitute back

u = x - 2

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

Simplify

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

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

Add a constant to the solution

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

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