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

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

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

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

- ln ∣ u + 1 ∣ + C

Solution steps

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

Simplify

\frac{1 + u}{- 2 u - 1 - u ^{2}} : - \frac{1}{u + 1}

= \int - \frac{1}{u + 1} d u

Take the constant out:

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

= - \int \frac{1}{u + 1} d u

Apply u-substitution

: \int \frac{1}{v} d v

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - ln ∣ v ∣

Substitute back

v = u + 1

= - ln ∣ u + 1 ∣

Add a constant to the solution

= - ln ∣ u + 1 ∣ + C

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y^{^{''}} + 5 y^{^{'}} + 6 y = 108 x^{2}

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