What is the integral of (2x)/((x^2+4)^2) ?

The integral of (2x)/((x^2+4)^2) is -1/(x^2+4)+C

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integral of (2x)/((x^2+4)^2)

\int \frac{2 x}{( x ^{2} + 4 ) ^{2}} d x

- \frac{1}{x ^{2} + 4} + C

Solution steps

\int \frac{2 x}{( x ^{2} + 4 ) ^{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}{( x ^{2} + 4 ) ^{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 = x^{2} + 4

= 2 \cdot \frac{1}{2} (- \frac{1}{x ^{2} + 4})

Simplify

2 \cdot \frac{1}{2} (- \frac{1}{x ^{2} + 4}) : - \frac{1}{x ^{2} + 4}

= - \frac{1}{x ^{2} + 4}

Add a constant to the solution

= - \frac{1}{x ^{2} + 4} + C

\frac{d y}{d x}, - x + y - x^{3} = y^{3} + 4 x^{2}

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

\int \frac{1}{n ( n + 1 )}

y^{^{''}} - 3 y^{^{'}} - 18 y = 0

x^{3} - 15 x^{2} + 50 x, - x^{3} + 15 x^{2} - 50 x

What is the integral of (2x)/((x^2+4)^2) ?
The integral of (2x)/((x^2+4)^2) is -1/(x^2+4)+C