What is the integral of (m^3)/((m^2+1)^3) ?

The integral of (m^3)/((m^2+1)^3) is (-2m^2-1)/(4(m^2+1)^2)+C

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integral of (m^3)/((m^2+1)^3)

\int \frac{m ^{3}}{( m ^{2} + 1 ) ^{3}} d m

\frac{- 2 m ^{2} - 1}{4 ( m ^{2} + 1 ) ^{2}} + C

Solution steps

\int \frac{m ^{3}}{( m ^{2} + 1 ) ^{3}} d m

Apply u-substitution

: \int \frac{u - 1}{2 u ^{3}} d u

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

Expand

\frac{u - 1}{u ^{3}} : \frac{1}{u ^{2}} - \frac{1}{u ^{3}}

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

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

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

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

Substitute back

u = m^{2} + 1

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

Simplify

\frac{1}{2} (- \frac{1}{m ^{2} + 1} - (- \frac{1}{2 ( m ^{2} + 1 ) ^{2}})) : \frac{- 2 m ^{2} - 1}{4 ( m ^{2} + 1 ) ^{2}}

= \frac{- 2 m ^{2} - 1}{4 ( m ^{2} + 1 ) ^{2}}

Add a constant to the solution

= \frac{- 2 m ^{2} - 1}{4 ( m ^{2} + 1 ) ^{2}} + C

(- 2, 5), (2, - 7)

x \to 2 - lim (\frac{- 2}{x ^{2} - 4})

\frac{\partial}{\partial y} (x - 5 y)

x \to 2 lim (7 - 3 x^{2} - x^{3})

\frac{d}{d x} (\frac{1}{4} e^{2 x^{2}})

What is the integral of (m^3)/((m^2+1)^3) ?
The integral of (m^3)/((m^2+1)^3) is (-2m^2-1)/(4(m^2+1)^2)+C