What is the integral of (x-5)/((x-5)^2+1) ?

The integral of (x-5)/((x-5)^2+1) is 1/2 ln|x^2-10x+26|+C

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integral of (x-5)/((x-5)^2+1)

\int \frac{x - 5}{( x - 5 ) ^{2} + 1} d x

\frac{1}{2} ln x^{2} - 10 x + 26 + C

Solution steps

\int \frac{x - 5}{( x - 5 ) ^{2} + 1} d x

Expand

\frac{x - 5}{( x - 5 ) ^{2} + 1} : \frac{x}{( x - 5 ) ^{2} + 1} - \frac{5}{( x - 5 ) ^{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{x}{( x - 5 ) ^{2} + 1} d x - \int \frac{5}{( x - 5 ) ^{2} + 1} d x

\int \frac{x}{( x - 5 ) ^{2} + 1} d x = \frac{1}{2} ln x^{2} - 10 x + 26 + 5 arctan (x - 5)

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

= \frac{1}{2} ln x^{2} - 10 x + 26 + 5 arctan (x - 5) - 5 arctan (x - 5)

Simplify

= \frac{1}{2} ln x^{2} - 10 x + 26

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 10 x + 26 + C

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What is the integral of (x-5)/((x-5)^2+1) ?
The integral of (x-5)/((x-5)^2+1) is 1/2 ln|x^2-10x+26|+C