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

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

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

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

- \frac{1}{2} ln 1 - 2 x - x^{2} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 u} d u

= \int - \frac{1}{2 u} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 - 2 x - x^{2}

= - \frac{1}{2} ln 1 - 2 x - x^{2}

Add a constant to the solution

= - \frac{1}{2} ln 1 - 2 x - x^{2} + C

690 (1 - (\frac{x}{345})^{2})

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2 y^{^{'}} = y^{3} \cdot cos (x)

\int e^{4 x - 5} d x

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