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

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

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

integral

\frac{( 1 - 2 v - v ^{2} )}{( 1 + v )}

2 ln ∣ v + 1 ∣ - \frac{1}{2} (v + 1)^{2} + C

Solution steps

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{1 + v} d v = ln ∣ 1 + v ∣

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

\int \frac{v ^{2}}{1 + v} d v = \frac{( 1 + v ) ^{2}}{2} - 2 (1 + v) + ln ∣ 1 + v ∣

= ln ∣ 1 + v ∣ - 2 (1 + v - ln ∣ 1 + v ∣) - (\frac{( 1 + v ) ^{2}}{2} - 2 (1 + v) + ln ∣ 1 + v ∣)

Simplify

ln ∣ 1 + v ∣ - 2 (1 + v - ln ∣ 1 + v ∣) - (\frac{( 1 + v ) ^{2}}{2} - 2 (1 + v) + ln ∣ 1 + v ∣) : 2 ln ∣ v + 1 ∣ - \frac{1}{2} (v + 1)^{2}

= 2 ln ∣ v + 1 ∣ - \frac{1}{2} (v + 1)^{2}

Add a constant to the solution

= 2 ln ∣ v + 1 ∣ - \frac{1}{2} (v + 1)^{2} + C

2 sin (x) - e^{x}

sin (x + 2 y)

(1 + x)^{0.5}

f (t) = e^{t} \cdot t^{\frac{1}{2}}

(x + 4 x - 5 i)^{3}

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