Evaluate the integral.
-$$\int \frac { \sqrt { 4 + 3 x } } { x } d x$$
A) $2 \sqrt { 4 + 3 x } + C$
B) $2 \sqrt { 4 + 3 x } - 2 \ln \left( \frac { \sqrt { 4 + 3 x } - 2 } { \sqrt { 4 + 3 x } + 2 } \right) + C$
C) $2 \sqrt { 4 + 3 x } + 2 \tan ^ { - 1 } \left( \frac { \sqrt { 4 + 3 x } } { 2 } \right) + C$
D) $2 \sqrt { 4 + 3 x } + 2 \ln \left( \frac { \sqrt { 4 + 3 x } - 2 } { \sqrt { 4 + 3 x } + 2 } \right) + C$

Question

Quizplus · Accepted Answer

This is a tricky integral that requires a substitution. Let $u = \sqrt{4+3x}$. Then $x = \frac{u^2-4}{3}$ and $du = \frac{3}{2\sqrt{4+3x}} dx$. Substituting into the original expression, we get:

$$\int \frac{\sqrt{4+3x}}{x} dx = \int \frac{u}{\frac{u^2-4}{3}} \cdot \frac{2\sqrt{4+3x}}{3} du = \frac{2}{3} \int \frac{u^2}{u^2-4} du$$

Now we use partial fractions to break up the fraction:

$$\frac{u^2}{u^2-4} = 1 + \frac{4}{u^2-4} = 1 + \frac{4}{(u-2)(u+2)}$$

Using the substitution $v = u-2$, we can turn this into

$$\frac{4}{(u-2)(u+2)} = \frac{2}{v} - \frac{2}{v+4}$$

Plugging back into the integral, we get

$$\int \frac{\sqrt{4+3x}}{x} dx = \frac{2}{3} \int \left(1 + \frac{2}{u-2} - \frac{2}{u+2}\right) du$$

Now we just need to integrate each term:

$$\int 1 du = u + C_1 = \sqrt{4+3x} + C_1$$

$$\int \frac{1}{u-2} du = \ln|u-2| + C_2 = \ln|\sqrt{4+3x}-2|+C_2$$

$$\int \frac{-1}{u+2} du = -\ln|u+2| + C_3 = -\ln|\sqrt{4+3x}+2|+C_3$$

Putting it all together, we get

$$\int \frac{\sqrt{4+3x}}{x} dx = \frac{2}{3} \left(\sqrt{4+3x} + \ln|\sqrt{4+3x}-2|-\ln|\sqrt{4+3x}+2|\right) + C$$

which simplifies to

$$\int \frac{\sqrt{4+3x}}{x} dx = 2\sqrt{4+3x} + \frac{2}{3} \ln\left|\frac{\sqrt{4+3x}-2}{\sqrt{4+3x}+2}\right| + C$$

Therefore, the answer is (D).

Evaluate the Integral ∫4+3xxdx\int \frac { \sqrt { 4 + 3 x } } { x } d x∫x4+3x​​dx

Evaluate the Integral $\int \frac { \sqrt { 4 + 3 x } } { x } d x$