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

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

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

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

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

Solution steps

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

Long division

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d x = x

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

= x - ln x^{2} + 1

Add a constant to the solution

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

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