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

f (t) = 7 + 10 x

\frac{d}{d x} (5 x ln (x))

\int \frac{1}{5 x - 3} d x

h \to 0 lim (\frac{( 2 ^{h} - 1 )}{h})

\frac{d}{d x} ((3 + 2 x) e^{- 2 x})

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