What is the integral of (3x)/((x^2+3)^2) ?

The integral of (3x)/((x^2+3)^2) is -3/(2(x^2+3))+C

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integral of (3x)/((x^2+3)^2)

\int \frac{3 x}{( x ^{2} + 3 ) ^{2}} d x

- \frac{3}{2 ( x ^{2} + 3 )} + C

Solution steps

\int \frac{3 x}{( x ^{2} + 3 ) ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{( x ^{2} + 3 ) ^{2}} d x

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 3 \cdot \frac{1}{2} (- \frac{1}{u})

Substitute back

u = x^{2} + 3

= 3 \cdot \frac{1}{2} (- \frac{1}{x ^{2} + 3})

Simplify

3 \cdot \frac{1}{2} (- \frac{1}{x ^{2} + 3}) : - \frac{3}{2 ( x ^{2} + 3 )}

= - \frac{3}{2 ( x ^{2} + 3 )}

Add a constant to the solution

= - \frac{3}{2 ( x ^{2} + 3 )} + C

f (x) = x + 2

x \to - 1 lim (x^{2} - 3 x + 1)

\frac{d}{d x} (\frac{3 x - 1}{( 2 - x ) ^{2}})

d x - 2 x y d y = 0

\frac{d}{d x} (\frac{( x - 1 ) ( x ^{2} + x + 1 )}{x ^{3}})

What is the integral of (3x)/((x^2+3)^2) ?
The integral of (3x)/((x^2+3)^2) is -3/(2(x^2+3))+C