2. $$ \\int \\frac { 1 } { 3 x + 12 } d x = $$\n(a) $$ - 3 \\ln | x + 4 | + c $$\n(b) $$ \\frac { 1 } { 3 }…

2. $$ \\int \\frac { 1 } { 3 x + 12 } d x = $$\n(a) $$ - 3 \\ln | x + 4 | + c $$\n(b) $$ \\frac { 1 } { 3 } \\ln | x + 4 | + c $$\n(c) $$ \\ln | x + 4 | + c $$\n(d) $$ 3 \\ln | x + 4 | + c $$

2. $$ \\int \\frac { 1 } { 3 x + 12 } d x = $$\n(a) $$ - 3 \\ln | x + 4 | + c $$\n(b) $$ \\frac { 1 } { 3 } \\ln | x + 4 | + c $$\n(c) $$ \\ln | x + 4 | + c $$\n(d) $$ 3 \\ln | x + 4 | + c $$

Answer

Explanation:

Step1: Simplify the integrand

First, factor out 3 from the denominator: $$\int\frac{1}{3x + 12}dx=\int\frac{1}{3(x + 4)}dx$$

Step2: Use the integral formula $\int\frac{1}{u}du=\ln|u|+C$

Let $u=x + 4$, then $du=dx$. $$\int\frac{1}{3(x + 4)}dx=\frac{1}{3}\int\frac{1}{u}du$$

Step3: Integrate

Using the formula $\int\frac{1}{u}du=\ln|u|+C$, we get: $$\frac{1}{3}\int\frac{1}{u}du=\frac{1}{3}\ln|u|+C$$

Step4: Substitute back

Substitute $u = x + 4$ back into the expression: $$\frac{1}{3}\ln|x + 4|+C$$

Answer:

B. $\frac{1}{3}\ln|x + 4|+C$