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

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

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

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

- \frac{5}{2 ( x ^{2} + 6 )} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 5 \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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 5 \cdot \frac{1}{2} (- \frac{1}{u})

Substitute back

u = x^{2} + 6

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

Simplify

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

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

Add a constant to the solution

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

\int 0 e^{x} d x

y^{^{''}} sin (x) cos (x) - 2 y^{^{'}} = 0

\int 6 e^{- 5 x} d x

x \to 0 lim (\frac{2}{sin ( x )})

y = lo g_{e} (x)

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