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

The integral of 1/(u(1+u^2)^2) is ln(u)-1/2 ln|1+u^2|+1/(2(1+u^2))+C

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integral of 1/(u(1+u^2)^2)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 ( v - 1 ) v ^{2}} d v

= \int \frac{1}{2 ( v - 1 ) v ^{2}} d v

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}{( v - 1 ) v ^{2}} d v

Take the partial fraction of

\frac{1}{( v - 1 ) v ^{2}} : \frac{1}{v - 1} - \frac{1}{v} - \frac{1}{v ^{2}}

= \frac{1}{2} \cdot \int \frac{1}{v - 1} - \frac{1}{v} - \frac{1}{v ^{2}} d v

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}{v - 1} d v - \int \frac{1}{v} d v - \int \frac{1}{v ^{2}} d v)

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

\int \frac{1}{v} d v = ln ∣ v ∣

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

= \frac{1}{2} (ln ∣ v - 1 ∣ - ln ∣ v ∣ - (- \frac{1}{v}))

Substitute back

v = 1 + u^{2}

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 1/(u(1+u^2)^2) ?
The integral of 1/(u(1+u^2)^2) is ln(u)-1/2 ln|1+u^2|+1/(2(1+u^2))+C