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

y = 14 x, y = x^{2}

4 x^{\frac{2}{3}}

\int 6 cos^{5} (x) sin^{4} (x) d x

x^{\frac{1}{2}}, x = 9

\frac{\partial}{\partial x} (b x e^{2 x y})

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