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: Tidy the integrand.
$\dfrac{\sqrt{x}}{x(x+1)} = \dfrac{\sqrt{x}}{x}\cdot\dfrac{1}{x+1} = \dfrac{1}{\sqrt{x}(x+1)}$.
Step 2: Choose a substitution.
Let $t = \sqrt{x}$, so $x = t^2$.
Step 3: Convert $dx$.
Differentiating, $dt = \dfrac{1}{2\sqrt{x}}\,dx$, hence $\dfrac{1}{\sqrt{x}}\,dx = 2\,dt$.
Step 4: Rewrite the whole integral in $t$.
$\displaystyle\int \dfrac{1}{x+1}\cdot\dfrac{1}{\sqrt{x}}\,dx = \int \dfrac{2}{t^2+1}\,dt$.
Step 5: Integrate the standard form.
$2\displaystyle\int \dfrac{dt}{t^2+1} = 2\tan^{-1}t + c$.
Step 6: Return to $x$ and compare.
$= 2\tan^{-1}(\sqrt{x}) + c$. Matching $k\tan^{-1}m + c$ gives $k = 2$, $m = \sqrt{x}$.
\[ \boxed{k = 2,\ m = \sqrt{x}} \]