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

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

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integral of x(x^2+1)^{-2}

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

= \frac{1}{2} (- \frac{1}{u})

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

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