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

The integral of u/((u^2+1)(1-2u)(1+2u)) is 1/10 ln|u^2+1|-1/10 ln|2u-1|-1/10 ln|2u+1|+C

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

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

\frac{1}{10} ln u^{2} + 1 - \frac{1}{10} ln ∣ 2 u - 1 ∣ - \frac{1}{10} ln ∣ 2 u + 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{u}{( u ^{2} + 1 ) ( 1 - 2 u ) ( 1 + 2 u )} : \frac{u}{5 ( u ^{2} + 1 )} - \frac{1}{5 ( 2 u - 1 )} - \frac{1}{5 ( 2 u + 1 )}

= \int \frac{u}{5 ( u ^{2} + 1 )} - \frac{1}{5 ( 2 u - 1 )} - \frac{1}{5 ( 2 u + 1 )} 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

= \int \frac{u}{5 ( u ^{2} + 1 )} d u - \int \frac{1}{5 ( 2 u - 1 )} d u - \int \frac{1}{5 ( 2 u + 1 )} d u

\int \frac{u}{5 ( u ^{2} + 1 )} d u = \frac{1}{10} ln u^{2} + 1

\int \frac{1}{5 ( 2 u - 1 )} d u = \frac{1}{10} ln ∣ 2 u - 1 ∣

\int \frac{1}{5 ( 2 u + 1 )} d u = \frac{1}{10} ln ∣ 2 u + 1 ∣

= \frac{1}{10} ln u^{2} + 1 - \frac{1}{10} ln ∣ 2 u - 1 ∣ - \frac{1}{10} ln ∣ 2 u + 1 ∣

Add a constant to the solution

= \frac{1}{10} ln u^{2} + 1 - \frac{1}{10} ln ∣ 2 u - 1 ∣ - \frac{1}{10} ln ∣ 2 u + 1 ∣ + C

\int \frac{7 x + 3}{x ^{2} - 2 x + 1} d x

\frac{e ^{2 s}}{s ^{3}}

\frac{\partial}{\partial x} (x y e^{s i n (π x y)})

x \to - 3 lim ((5 - x) \frac{4}{3})

\int \frac{5}{1 + 9 x ^{2}} d x

What is the integral of u/((u^2+1)(1-2u)(1+2u)) ?
The integral of u/((u^2+1)(1-2u)(1+2u)) is 1/10 ln|u^2+1|-1/10 ln|2u-1|-1/10 ln|2u+1|+C