What is the integral of (6x)/((x^2+5)^2) ?

The integral of (6x)/((x^2+5)^2) is -3/(x^2+5)+C

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integral of (6x)/((x^2+5)^2)

\int \frac{6 x}{( x ^{2} + 5 ) ^{2}} d x

- \frac{3}{x ^{2} + 5} + C

Solution steps

\int \frac{6 x}{( x ^{2} + 5 ) ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{( x ^{2} + 5 ) ^{2}} d x

Apply u-substitution

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

= 6 \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

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 6 \cdot \frac{1}{2} (- \frac{1}{u})

Substitute back

u = x^{2} + 5

= 6 \cdot \frac{1}{2} (- \frac{1}{x ^{2} + 5})

Simplify

6 \cdot \frac{1}{2} (- \frac{1}{x ^{2} + 5}) : - \frac{3}{x ^{2} + 5}

= - \frac{3}{x ^{2} + 5}

Add a constant to the solution

= - \frac{3}{x ^{2} + 5} + C

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 10, y^{^{'}} (0) = 0

\int_{0}^{\frac{π}{4}} sin (2 x) d x

\frac{\partial}{\partial x} (\frac{( 2 x - y )}{x y})

\frac{d}{d x} (a ln (x))

f (x) 0

What is the integral of (6x)/((x^2+5)^2) ?
The integral of (6x)/((x^2+5)^2) is -3/(x^2+5)+C