What is the integral of (x+3)/(x^2-2x+5) ?

The integral of (x+3)/(x^2-2x+5) is 1/2 ln| 1/4 (x-1)^2+1|+2arctan(1/2 (x-1))+C

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integral of (x+3)/(x^2-2x+5)

\int \frac{x + 3}{x ^{2} - 2 x + 5} d x

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

Solution steps

\int \frac{x + 3}{x ^{2} - 2 x + 5} d x

Complete the square

x^{2} - 2 x + 5 : (x - 1)^{2} + 4

= \int \frac{x + 3}{( x - 1 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{u + 4}{u ^{2} + 4} d u

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

Apply Integral Substitution

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

Expand

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

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

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (x+3)/(x^2-2x+5) ?
The integral of (x+3)/(x^2-2x+5) is 1/2 ln| 1/4 (x-1)^2+1|+2arctan(1/2 (x-1))+C