Question

$\int\frac{1}{x+\sqrt{x}}dx=$

• A. $\log|1+\sqrt{x}|+C$
• B. $2\log|1-\sqrt{x}|+C$
• C. $\log|1-\sqrt{x}|+C$
• D. $2\log|1+\sqrt{x}|+C$
• E. $2\log|x+\sqrt{x}|+C$

Hint:

Algebra Tip: You can also solve this by letting $x = t^2$. Then $dx = 2t dt$, and the integral becomes $\int \frac{2t dt}{t^2 + t} = \int \frac{2 dt}{t + 1} = 2 \log|t + 1| + C$. Both methods are excellent!

Answer 1

The Correct Option is D

Concept:
Integrals containing a mix of $x$ and $\sqrt{x}$ can be solved by factoring out the lowest power of $x$ (which is $\sqrt{x}$) in the denominator. This sets up a perfect scenario for u-substitution.

Step 1: Factor the denominator.
The integral is given as: $$I = \int \frac{1}{x + \sqrt{x}} dx$$ Recognize that $x = (\sqrt{x})^2$. Factor out $\sqrt{x}$ from both terms in the denominator: $$I = \int \frac{1}{\sqrt{x}(\sqrt{x} + 1)} dx$$

Step 2: Choose the substitution variable u.
Let $u$ equal the expression inside the parentheses: $$u = \sqrt{x} + 1$$

Step 3: Differentiate u to find du.
Take the derivative of $u$ with respect to $x$: $$\frac{du}{dx} = \frac{1}{2\sqrt{x}}$$ Multiply both sides by 2 to isolate the remaining part of the integrand: $$2 du = \frac{1}{\sqrt{x}} dx$$

Step 4: Substitute into the integral and evaluate.
Replace the components in the factored integral with $u$ and $du$: $$I = \int \frac{1}{u} \cdot (2 du) = 2 \int \frac{1}{u} du$$ Integrate using the natural logarithm rule: $$I = 2 \log|u| + C$$

Step 5: Substitute x back into the equation.
Replace $u$ with its original definition: $$I = 2 \log|\sqrt{x} + 1| + C$$ Since addition is commutative, this is identical to $2 \log|1 + \sqrt{x}| + C$. Hence the correct answer is (D) $2\log|1+\sqrt{x|+C$}.