Question

If $\int \frac{\sqrt{x}}{x(x+1)} \, dx = k \tan^{-1} m + c$ (where $c$ is a constant of integration), then

• A. $k = 1, \ m = x$
• B. $k = 2, \ m = x$
• C. $k = 1, \ m = \sqrt{x}$
• D. $k = 2, \ m = \sqrt{x}$

Hint:

Whenever you see an integral featuring $\sqrt{x}$ mixed with a standard polynomial, substituting $x = t^2$ is an excellent time-saver because it cleans up the fractional powers immediately!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks us to evaluate the indefinite integral $I = \int \frac{\sqrt{x}}{x(x+1)} \, dx$ and compare it to the given pattern $k \tan^{-1} m + c$ to identify the values of the parameter $k$ and the function $m$.

Step 2: Key Formula or Approach:
We can simplify the integrand by algebraically reducing the power of $x$ in the numerator or by substituting a new variable for the radical term, $t = \sqrt{x}$.

Step 3: Detailed Explanation:
Let's first simplify the integrand directly:
$$I = \int \frac{\sqrt{x}}{x(x+1)} \, dx = \int \frac{1}{\sqrt{x}(x+1)} \, dx$$ Now, substitute $t = \sqrt{x} \implies x = t^2$. Differentiating both sides gives:
$$dt = \frac{1}{2\sqrt{x}} \, dx \implies \frac{1}{\sqrt{x}} \, dx = 2 \, dt$$ Substitute these variables into the simplified integral:
$$I = \int \frac{2}{t^2 + 1} \, dt = 2 \int \frac{1}{t^2 + 1} \, dt$$ Apply the standard standard integration formula $\int \frac{1}{t^2+1} \, dt = \tan^{-1} t$:
$$I = 2 \tan^{-1} t + c$$ Substitute back $t = \sqrt{x}$ to get the final answer in terms of $x$:
$$I = 2 \tan^{-1}(\sqrt{x}) + c$$ Comparing this with the given target form $k \tan^{-1} m + c$, we find:
$k = 2$
$m = \sqrt{x}$
This corresponds exactly to option (D).

Step 4: Final Answer:
The values are $k = 2$ and $m = \sqrt{x}$, which corresponds to option (D).