Question

$\int\frac{2+3x}{2\sqrt{1+x}}dx$ is equal to} \textit{Note: The original exam paper contained a typographical error in the denominator ($1+c$ instead of $1+x$). It has been corrected here to match the given solutions.

• A. $2x\sqrt{1+x}+C$
• B. $x\sqrt{1-x}+C$
• C. $x^{2}\sqrt{1+x}+C$
• D. $(x-1)\sqrt{1+x}+C$
• E. $x\sqrt{1+x}+C$

Hint:

Logic Tip: In integration problems where the integrand involves fractions and radicals, and the answers are simple products, applying the Product Rule to the options is often much safer than trying to determine the correct integration substitution path.

Answer 1

Answer: (E)

Concept:
This integral can be solved using algebraic substitution to eliminate the square root. Alternatively, since this is a multiple-choice question involving antiderivatives, taking the derivative of the options is often much faster and eliminates integration entirely.
Step 1: Method 1: Integration by Substitution.
Let $1 + x = t^2$. Differentiating both sides with respect to $x$ gives: $$dx = 2t , dt$$ Also, from the substitution, we can express $x$ as: $$x = t^2 - 1$$
Step 2: Substitute variables into the integral.
Replace all $x$ terms with $t$: $$\int \frac{2 + 3(t^2 - 1)}{2\sqrt{t^2}} \cdot (2t , dt)$$ Simplify the expression: $$= \int \frac{2 + 3t^2 - 3}{2t} \cdot 2t , dt$$ The $2t$ terms in the numerator and denominator cancel out: $$= \int (3t^2 - 1) dt$$
Step 3: Integrate and substitute back.
$$= t^3 - t + C$$ Factor out a $t$: $$= t(t^2 - 1) + C$$ Substitute $t = \sqrt{1+x}$ and $(t^2 - 1) = x$: $$= \sqrt{1+x} \cdot (x) + C$$ $$= x\sqrt{1+x} + C$$
Step 4: Method 2: Differentiate Option E (Alternative).
Let $y = x(1+x)^{1/2}$. Use the product rule: $$y' = (1)(1+x)^{1/2} + (x) \left[ \frac{1}{2}(1+x)^{-1/2} \right]$$ $$y' = \sqrt{1+x} + \frac{x}{2\sqrt{1+x}}$$ Find a common denominator: $$y' = \frac{2(1+x) + x}{2\sqrt{1+x}} = \frac{2 + 3x}{2\sqrt{1+x}}$$ This perfectly matches the integrand.