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

The integral of u/(1+u-2u^2) is 2(-1/8 ln|-16u^2+8u+8|+1/24 (ln| 2/3+4/3 u|-ln|-4/3+4/3 u|))+C

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

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

2 (- \frac{1}{8} ln - 16 u^{2} + 8 u + 8 + \frac{1}{24} (ln \frac{2}{3} + \frac{4}{3} u - ln - \frac{4}{3} + \frac{4}{3} u)) + C

Solution steps

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

Complete the square

1 + u - 2 u^{2} : - 2 (u - \frac{1}{4})^{2} + \frac{9}{8}

= \int \frac{u}{- 2 ( u - \frac{1}{4} ) ^{2} + \frac{9}{8}} d u

Apply u-substitution

: \int \frac{2 ( 4 v + 1 )}{- 16 v ^{2} + 9} d v

= \int \frac{2 ( 4 v + 1 )}{- 16 v ^{2} + 9} d v

Take the constant out:

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

= 2 \cdot \int \frac{4 v + 1}{- 16 v ^{2} + 9} d v

Expand

\frac{4 v + 1}{- 16 v ^{2} + 9} : \frac{4 v}{- 16 v ^{2} + 9} + \frac{1}{- 16 v ^{2} + 9}

Apply the Sum Rule:

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

= 2 (\int \frac{4 v}{- 16 v ^{2} + 9} d v + \int \frac{1}{- 16 v ^{2} + 9} d v)

\int \frac{4 v}{- 16 v ^{2} + 9} d v = - \frac{1}{8} ln - 16 v^{2} + 9

\int \frac{1}{- 16 v ^{2} + 9} d v = \frac{1}{24} (ln \frac{4}{3} v + 1 - ln \frac{4}{3} v - 1)

= 2 (- \frac{1}{8} ln - 16 v^{2} + 9 + \frac{1}{24} (ln \frac{4}{3} v + 1 - ln \frac{4}{3} v - 1))

Substitute back

v = u - \frac{1}{4}

= 2 (- \frac{1}{8} ln - 16 (u - \frac{1}{4})^{2} + 9 + \frac{1}{24} (ln \frac{4}{3} (u - \frac{1}{4}) + 1 - ln \frac{4}{3} (u - \frac{1}{4}) - 1))

Simplify

2 (- \frac{1}{8} ln - 16 (u - \frac{1}{4})^{2} + 9 + \frac{1}{24} (ln \frac{4}{3} (u - \frac{1}{4}) + 1 - ln \frac{4}{3} (u - \frac{1}{4}) - 1)) : 2 (- \frac{1}{8} ln - 16 u^{2} + 8 u + 8 + \frac{1}{24} (ln \frac{2}{3} + \frac{4}{3} u - ln - \frac{4}{3} + \frac{4}{3} u))

= 2 (- \frac{1}{8} ln - 16 u^{2} + 8 u + 8 + \frac{1}{24} (ln \frac{2}{3} + \frac{4}{3} u - ln - \frac{4}{3} + \frac{4}{3} u))

Add a constant to the solution

= 2 (- \frac{1}{8} ln - 16 u^{2} + 8 u + 8 + \frac{1}{24} (ln \frac{2}{3} + \frac{4}{3} u - ln - \frac{4}{3} + \frac{4}{3} u)) + C

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What is the integral of u/(1+u-2u^2) ?
The integral of u/(1+u-2u^2) is 2(-1/8 ln|-16u^2+8u+8|+1/24 (ln| 2/3+4/3 u|-ln|-4/3+4/3 u|))+C