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

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

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

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

ln x^{2} - 2 x + 5 + arctan (\frac{1}{2} (x - 1)) + C

Solution steps

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

: \int \frac{u + 1}{u ^{2} + 4} d u

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

Expand

\frac{u + 1}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} + \frac{1}{u ^{2} + 4}

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{u}{u ^{2} + 4} d u + \int \frac{1}{u ^{2} + 4} d u)

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

\int \frac{1}{u ^{2} + 4} d u = \frac{1}{2} arctan (\frac{u}{2})

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

Substitute back

u = x - 1

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

Simplify

2 (\frac{1}{2} ln (x - 1)^{2} + 4 + \frac{1}{2} arctan (\frac{x - 1}{2})) : ln x^{2} - 2 x + 5 + arctan (\frac{1}{2} (x - 1))

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

Add a constant to the solution

= ln x^{2} - 2 x + 5 + arctan (\frac{1}{2} (x - 1)) + C

x \to 0 lim (\frac{( x + 64 ) ^{\frac{1}{3}} - 4}{x})

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

\int_{1}^{\infty} x^{- \frac{1}{5}} d x

x \to 1 + lim (\frac{x + 1}{x ^{2} - 1})

\int 2 sin (x) + 3 cos (x) d x

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