What is the integral of (2-x)/(1+4x-x^2) ?

The integral of (2-x)/(1+4x-x^2) is 1/2 ln|1+4x-x^2|+C

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integral of (2-x)/(1+4x-x^2)

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

\frac{1}{2} ln 1 + 4 x - x^{2} + C

Solution steps

\int \frac{2 - x}{1 + 4 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 + 4 x - x^{2}

= \frac{1}{2} ln 1 + 4 x - x^{2}

Add a constant to the solution

= \frac{1}{2} ln 1 + 4 x - x^{2} + C

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What is the integral of (2-x)/(1+4x-x^2) ?
The integral of (2-x)/(1+4x-x^2) is 1/2 ln|1+4x-x^2|+C