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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

\int x e^{3} d x

- \frac{48 x}{( x ^{2} + 12 ) ^{2}}

\int_{0}^{9} \frac{1}{x ^{2} - 6 x + 5} d x

\frac{1}{1 + x ^{2}}, \frac{x ^{2}}{2}

\frac{\partial}{\partial x} (y arctan (\frac{x}{z}))

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