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

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

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

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

arctan (x) - x + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

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

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

\int 1 d x = x

= - (- arctan (x) + x)

Simplify

= arctan (x) - x

Add a constant to the solution

= arctan (x) - x + C

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