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

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

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

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

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

Solution steps

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

Factor

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

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

Apply Integration By Parts

: - \frac{ln ( v )}{v + 1} - \int - \frac{1}{v ( v + 1 )} d v

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

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

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

Simplify

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

Add a constant to the solution

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

x^{4} + 9 x^{2} - x (1.9)

f (x) = x^{2} + 2, at x = - 1

\frac{d}{d t} (\frac{1}{1 + 4 t ^{2}})

\frac{\partial}{\partial x} (\frac{x e ^{y}}{x + y})

x \to 1 lim (2^{\frac{1}{x - 1}})

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