What is the integral of 1/(5+2x-2x^2) ?

The integral of 1/(5+2x-2x^2) is 1/(2sqrt(11))(ln|(2x-1)/(sqrt(11))+1|-ln|(2x-1)/(sqrt(11))-1|)+C

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integral of 1/(5+2x-2x^2)

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

\frac{1}{2 11} (ln \frac{2 x - 1}{11} + 1 - ln \frac{2 x - 1}{11} - 1) + C

Solution steps

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

Complete the square

5 + 2 x - 2 x^{2} : - 2 (x - \frac{1}{2})^{2} + \frac{11}{2}

= \int \frac{1}{- 2 ( x - \frac{1}{2} ) ^{2} + \frac{11}{2}} d x

Apply u-substitution

= \int \frac{2}{- 4 u ^{2} + 11} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{- 4 u ^{2} + 11} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{2 11 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2 11} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

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

= 2 \cdot \frac{1}{2 11} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= 2 \cdot \frac{1}{2 11} \frac{ln \frac{2}{11} ( x - \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{11} ( x - \frac{1}{2} ) - 1}{2}

Simplify

2 \cdot \frac{1}{2 11} \frac{ln \frac{2}{11} ( x - \frac{1}{2} ) + 1}{2} - \frac{ln \frac{2}{11} ( x - \frac{1}{2} ) - 1}{2} : \frac{1}{2 11} (ln \frac{2 x - 1}{11} + 1 - ln \frac{2 x - 1}{11} - 1)

= \frac{1}{2 11} (ln \frac{2 x - 1}{11} + 1 - ln \frac{2 x - 1}{11} - 1)

Add a constant to the solution

= \frac{1}{2 11} (ln \frac{2 x - 1}{11} + 1 - ln \frac{2 x - 1}{11} - 1) + C

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What is the integral of 1/(5+2x-2x^2) ?
The integral of 1/(5+2x-2x^2) is 1/(2sqrt(11))(ln|(2x-1)/(sqrt(11))+1|-ln|(2x-1)/(sqrt(11))-1|)+C