What is the integral of (w^2)/((w^2+7)^2) ?

The integral of (w^2)/((w^2+7)^2) is 1/(2sqrt(7))arctan(w/(sqrt(7)))-w/(2(7+w^2))+C

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integral of (w^2)/((w^2+7)^2)

\int \frac{w ^{2}}{( w ^{2} + 7 ) ^{2}} d w

\frac{1}{2 7} arctan (\frac{w}{7}) - \frac{w}{2 ( 7 + w ^{2} )} + C

Solution steps

\int \frac{w ^{2}}{( w ^{2} + 7 ) ^{2}} d w

Apply Integral Substitution

= \int \frac{u ^{2}}{7 ( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{u ^{2}}{( u ^{2} + 1 ) ^{2}} d u

Take the partial fraction of

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

= \frac{1}{7} \cdot \int \frac{1}{u ^{2} + 1} - \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{7} (\int \frac{1}{u ^{2} + 1} d u - \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u)

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int \frac{1}{( u ^{2} + 1 ) ^{2}} d u = \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}})

= \frac{1}{7} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

Substitute back

u = \frac{w}{7}

= \frac{1}{7} arctan (\frac{w}{7}) - \frac{1}{2} arctan (\frac{w}{7}) + \frac{\frac{w}{7}}{1 + ( \frac{w}{7} ) ^{2}}

Simplify

\frac{1}{7} arctan (\frac{w}{7}) - \frac{1}{2} arctan (\frac{w}{7}) + \frac{\frac{w}{7}}{1 + ( \frac{w}{7} ) ^{2}} : \frac{1}{2 7} arctan (\frac{w}{7}) - \frac{w}{2 ( 7 + w ^{2} )}

= \frac{1}{2 7} arctan (\frac{w}{7}) - \frac{w}{2 ( 7 + w ^{2} )}

Add a constant to the solution

= \frac{1}{2 7} arctan (\frac{w}{7}) - \frac{w}{2 ( 7 + w ^{2} )} + C

y^{^{'}} = - \frac{2 x}{y}, y (1) = 10

y = (arctan (7 x))^{2}

x \to 8 lim (x^{- 1})

\frac{d}{d u} (e^{u} cos (v))

2 sin (3 x)

What is the integral of (w^2)/((w^2+7)^2) ?
The integral of (w^2)/((w^2+7)^2) is 1/(2sqrt(7))arctan(w/(sqrt(7)))-w/(2(7+w^2))+C