Find the indefinite integral. $\int v \ln ( v + 2 ) d v$
A) $\left( \frac { v ^ { 2 } - 4 } { 2 } \right) \ln ( v + 2 ) - \frac { v ^ { 2 } - 4 v } { 4 } + C$
B) $\left( \frac { v ^ { 2 } - 4 } { 2 } \right) \ln ( v + 2 ) + \frac { v ^ { 2 } - 4 v } { 4 } + C$
C) $\left( \frac { v ^ { 2 } + 4 } { 2 } \right) \ln ( v + 2 ) - \frac { v ^ { 2 } + 2 v } { 4 } + C$
D) $\left( \frac { v ^ { 2 } - 4 } { 2 } \right) \ln ( v + 2 ) - \frac { v ^ { 2 } - 4 v } { 2 } + C$
E) $\left( \frac { v ^ { 2 } - 4 } { 2 } \right) \ln ( v + 2 ) + \frac { v ^ { 2 } - 2 v } { 4 } + C$

Question

Quizplus · Accepted Answer

This integral can be solved using integration by parts, where $u = \ln(v + 2)$ and $dv = v dv$. The derivative of $u$ is $du = \frac{1}{v + 2} dv$, and the integral of $dv$ is $v^2/2$. Applying the integration by parts formula, $\int u dv = uv - \int v du$, and simplifying leads to the expression given in option A.

Find the Indefinite Integral ∫vln⁡(v+2)dv\int v \ln ( v + 2 ) d v∫vln(v+2)dv

Find the Indefinite Integral $\int v \ln ( v + 2 ) d v$