What is the integral of (3x-2)/(x^2-6x+10) ?

The integral of (3x-2)/(x^2-6x+10) is 3/2 ln|x^2-6x+10|+7arctan(x-3)+C

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integral of (3x-2)/(x^2-6x+10)

\int \frac{3 x - 2}{x ^{2} - 6 x + 10} d x

\frac{3}{2} ln x^{2} - 6 x + 10 + 7 arctan (x - 3) + C

Solution steps

\int \frac{3 x - 2}{x ^{2} - 6 x + 10} d x

Complete the square

x^{2} - 6 x + 10 : (x - 3)^{2} + 1

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

Apply u-substitution

: \int \frac{3 u + 7}{u ^{2} + 1} d u

= \int \frac{3 u + 7}{u ^{2} + 1} d u

Expand

\frac{3 u + 7}{u ^{2} + 1} : \frac{3 u}{u ^{2} + 1} + \frac{7}{u ^{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

= \int \frac{3 u}{u ^{2} + 1} d u + \int \frac{7}{u ^{2} + 1} d u

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

\int \frac{7}{u ^{2} + 1} d u = 7 arctan (u)

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

Substitute back

u = x - 3

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

Simplify

\frac{3}{2} ln (x - 3)^{2} + 1 + 7 arctan (x - 3) : \frac{3}{2} ln x^{2} - 6 x + 10 + 7 arctan (x - 3)

= \frac{3}{2} ln x^{2} - 6 x + 10 + 7 arctan (x - 3)

Add a constant to the solution

= \frac{3}{2} ln x^{2} - 6 x + 10 + 7 arctan (x - 3) + C

\frac{d}{d x} (1 0^{x l n (x) - x})

\int 2 e^{x} sin (x) d x

x \to 0 lim ((- 3 x + 1)^{\frac{5}{x}})

\int (x + 2)^{13} d x

f (x) = x^{2} - 3 x

What is the integral of (3x-2)/(x^2-6x+10) ?
The integral of (3x-2)/(x^2-6x+10) is 3/2 ln|x^2-6x+10|+7arctan(x-3)+C