What is the integral of s/((s^2+16)^2) ?

The integral of s/((s^2+16)^2) is -1/(2(s^2+16))+C

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integral of s/((s^2+16)^2)

\int \frac{s}{( s ^{2} + 16 ) ^{2}} d s

- \frac{1}{2 ( s ^{2} + 16 )} + C

Solution steps

\int \frac{s}{( s ^{2} + 16 ) ^{2}} d s

Apply u-substitution

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

= \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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = s^{2} + 16

= \frac{1}{2} (- \frac{1}{s ^{2} + 16})

Simplify

\frac{1}{2} (- \frac{1}{s ^{2} + 16}) : - \frac{1}{2 ( s ^{2} + 16 )}

= - \frac{1}{2 ( s ^{2} + 16 )}

Add a constant to the solution

= - \frac{1}{2 ( s ^{2} + 16 )} + C

\frac{d}{d x} (100 (y - x^{2})^{2} + (1 - x)^{2})

\frac{d}{d x} (\frac{24 x}{2 x + 3})

x \to 0 lim (\frac{x}{- x})

t^{\frac{3}{2}}

\int (x^{2} + \frac{1}{4} y^{2} + 3) d x

What is the integral of s/((s^2+16)^2) ?
The integral of s/((s^2+16)^2) is -1/(2(s^2+16))+C