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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{1}{s ^{\frac{3}{2}}}

6 y^{^{''}} + 31 y = 0, y (0) = 6, y^{^{'}} (0) = 2

\frac{d}{d x} (\frac{1}{y ^{2} + x})

y = \frac{1}{x ^{2}}, (3, \frac{1}{9})

\frac{d y}{d x} - \frac{9}{x} y = \frac{y ^{3}}{x ^{13}}

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