# Simple integral: int {-3x+5}/{x^2-2x+5}dx=?

## $\int \frac{3 {x}^{2} + 5 x + 50}{{x}^{4} + {x}^{2} + 10 x} \mathrm{dx} = \ldots = \int \left(\frac{5}{x} - \frac{2}{x + 2} + \frac{- 3 x + 5}{{x}^{2} - 2 x + 5}\right) \mathrm{dx} = 5 \ln | x | - 2 \ln | x + 2 | + S O M E T H I N G + c$ whtat is the "something" (int {-3x+5}/{x^2-2x+5}dx=?)?

Jan 28, 2018

$\int \frac{- 3 x + 5}{{x}^{2} - 2 x + 5} \cdot \mathrm{dx}$

$= \arctan \left(\frac{x - 1}{2}\right) - \frac{3}{2} \ln \left({x}^{2} - 2 x + 5\right)$

#### Explanation:

$\int \frac{- 3 x + 5}{{x}^{2} - 2 x + 5} \cdot \mathrm{dx}$

=$- \int \frac{3 x - 5}{{x}^{2} - 2 x + 5} \cdot \mathrm{dx}$

=$- \int \frac{3 x - 3 - 2}{{x}^{2} - 2 x + 5} \cdot \mathrm{dx}$

=$- \int \frac{3 x - 3}{{x}^{2} - 2 x + 5} \cdot \mathrm{dx}$+$\int \frac{2}{{x}^{2} - 2 x + 5} \cdot \mathrm{dx}$

=$\int \frac{2}{{\left(x - 1\right)}^{2} + 4} \cdot \mathrm{dx}$-$\frac{3}{2} \int \frac{2 x - 2}{{x}^{2} - 2 x + 5}$

=$\arctan \left(\frac{x - 1}{2}\right) - \frac{3}{2} \ln \left({x}^{2} - 2 x + 5\right)$

Jan 28, 2018

$= - \frac{3}{2} \ln \left({x}^{2} - 2 x + 5\right) + {\tan}^{-} 1 \left(\frac{x - 1}{2}\right) + C$

#### Explanation:

$\int \frac{- 3 x + 5}{{x}^{2} - 2 x + 5} \mathrm{dx}$

$= \int \frac{- 3 x + 5 - 2 + 2}{{x}^{2} - 2 x + 5} \mathrm{dx}$

$= \int \frac{- 3 x + 3}{{x}^{2} - 2 x + 5} + \frac{2}{{x}^{2} - 2 x + 5} \mathrm{dx}$

$= - \int \frac{3 x - 3}{{x}^{2} - 2 x + 5} \mathrm{dx} + \int \frac{2}{{x}^{2} - 2 x + 5} \mathrm{dx}$

For:

$- \int \frac{3 x - 3}{{x}^{2} - 2 x + 5} \mathrm{dx}$

Use the substitution:

$u = {x}^{2} - 2 x + 5$

$\implies \mathrm{du} = 2 x - 2 \mathrm{dx} \implies \frac{3}{2} \mathrm{du} = 3 x - 3 \mathrm{dx}$

$\therefore - \int \frac{3 x - 3}{{x}^{2} - 2 x + 5} \mathrm{dx} = - \int \frac{\frac{3}{2}}{u} \mathrm{du} = - \frac{3}{2} \ln \left(u\right) + C$

Reverse the substitution:

$- \frac{3}{2} \ln \left({x}^{2} - 2 x + 5\right) + C$

Now for the other integral:

$\int \frac{2}{{x}^{2} - 2 x + 5} \mathrm{dx}$

Write the denominator in completed square form:

${x}^{2} - 2 x + 5 = {\left(x - 1\right)}^{2} - {\left(- 1\right)}^{2} + 5 = {\left(x - 1\right)}^{2} + 4$

So:

$\int \frac{2}{{x}^{2} - 2 x + 5} \mathrm{dx} = 2 \int \frac{\mathrm{dx}}{{\left(x - 1\right)}^{2} + 4}$

Now substitute:

$2 u = \left(x - 1\right)$

$\implies \mathrm{du} = 2 \mathrm{dx}$ So:

$2 \int \frac{\mathrm{dx}}{{\left(x - 1\right)}^{2} + 4} = 2 \int \frac{2}{4 {u}^{2} + 4} \mathrm{du} = \frac{4}{4} \int \frac{1}{{u}^{2} + 1} \mathrm{du}$

Which we recognize will simply integrate to inverse tangent giving us:

$= {\tan}^{-} 1 \left(u\right) + C '$

Reverse the substitution:

$= {\tan}^{-} 1 \left(\frac{x - 1}{2}\right) + C '$

Hence, the "something" is:

$\int \frac{- 3 x + 5}{{x}^{2} - 2 x + 5} \mathrm{dx}$

$= - \int \frac{3 x - 3}{{x}^{2} - 2 x + 5} \mathrm{dx} + \int \frac{2}{{x}^{2} - 2 x + 5} \mathrm{dx}$

$= - \frac{3}{2} \ln \left({x}^{2} - 2 x + 5\right) + {\tan}^{-} 1 \left(\frac{x - 1}{2}\right) + C$