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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= - 2 \cdot \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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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

x^{4} - 3 x^{2} + 1, 1

y = e^{x} (x + x x)

x \to - 2 lim (4 x^{2} - 3)

n = - 4 \sum \infty (- \frac{4}{5})^{n}

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