What is the integral of x/(3-2x-x^2) ?

The integral of x/(3-2x-x^2) is

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integral of x/(3-2x-x^2)

\int \frac{x}{3 - 2 x - x ^{2}} d x

\frac{1}{4} x (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1) - \frac{1}{4} (ln \frac{x + 1}{2} + 1 (x + 1) - ln \frac{x + 1}{2} - 1 (x + 1) + 2 ln ∣ x + 3 ∣ + 2 ln ∣ x - 1 ∣ - 4) + C

Solution steps

\int \frac{x}{3 - 2 x - x ^{2}} d x

Complete the square

3 - 2 x - x^{2} : - (x + 1)^{2} + 4

= \int \frac{x}{- ( x + 1 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{u - 1}{- u ^{2} + 4} d u

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

Apply Integration By Parts

: \frac{1}{4} (u - 1) (ln \frac{u}{2} + 1 - ln \frac{u}{2} - 1) - \int \frac{1}{4} (ln \frac{u}{2} + 1 - ln \frac{u}{2} - 1) d u

= \frac{1}{4} (u - 1) (ln \frac{u}{2} + 1 - ln \frac{u}{2} - 1) - \int \frac{1}{4} (ln \frac{u}{2} + 1 - ln \frac{u}{2} - 1) d u

\int \frac{1}{4} (ln \frac{u}{2} + 1 - ln \frac{u}{2} - 1) d u = \frac{1}{4} (u ln \frac{u}{2} + 1 - u ln \frac{u}{2} - 1 + 2 ln ∣ u + 2 ∣ + 2 ln ∣ u - 2 ∣ - 4)

= \frac{1}{4} (u - 1) (ln \frac{u}{2} + 1 - ln \frac{u}{2} - 1) - \frac{1}{4} (u ln \frac{u}{2} + 1 - u ln \frac{u}{2} - 1 + 2 ln ∣ u + 2 ∣ + 2 ln ∣ u - 2 ∣ - 4)

Substitute back

u = x + 1

= \frac{1}{4} (x + 1 - 1) (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1) - \frac{1}{4} ((x + 1) ln \frac{x + 1}{2} + 1 - (x + 1) ln \frac{x + 1}{2} - 1 + 2 ln ∣ x + 1 + 2 ∣ + 2 ln ∣ x + 1 - 2 ∣ - 4)

Simplify

\frac{1}{4} (x + 1 - 1) (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1) - \frac{1}{4} ((x + 1) ln \frac{x + 1}{2} + 1 - (x + 1) ln \frac{x + 1}{2} - 1 + 2 ln ∣ x + 1 + 2 ∣ + 2 ln ∣ x + 1 - 2 ∣ - 4) : \frac{1}{4} x (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1) - \frac{1}{4} (ln \frac{x + 1}{2} + 1 (x + 1) - ln \frac{x + 1}{2} - 1 (x + 1) + 2 ln ∣ x + 3 ∣ + 2 ln ∣ x - 1 ∣ - 4)

= \frac{1}{4} x (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1) - \frac{1}{4} (ln \frac{x + 1}{2} + 1 (x + 1) - ln \frac{x + 1}{2} - 1 (x + 1) + 2 ln ∣ x + 3 ∣ + 2 ln ∣ x - 1 ∣ - 4)

Add a constant to the solution

= \frac{1}{4} x (ln \frac{x + 1}{2} + 1 - ln \frac{x + 1}{2} - 1) - \frac{1}{4} (ln \frac{x + 1}{2} + 1 (x + 1) - ln \frac{x + 1}{2} - 1 (x + 1) + 2 ln ∣ x + 3 ∣ + 2 ln ∣ x - 1 ∣ - 4) + C

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