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

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

Popular >

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

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

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

Solution steps

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

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 = 2 - 2 p - 4 p^{2}

= - \frac{1}{2} ln 2 - 2 p - 4 p^{2}

Add a constant to the solution

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

\int_{0}^{4} (\frac{1}{4} x + (x + 1)) d x

\frac{d}{d x} (e^{- \frac{a}{x}})

x \to 3 lim (\frac{4 x - 12}{x - 3})

\int \frac{sin ( x )}{x} d x

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

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