Question

$\int x(1+2\log x)dx$ is equal to

• A. $x\log x+C$
• B. $x^{2}\log x+C$
• C. $x^{3}\log x+C$
• D. $2x\log x+C$
• E. $3x\log x+C$

Hint:

Logic Tip: In calculus exams, "$\log x$" almost exclusively refers to the natural logarithm "$\ln x$". Also, differentiating options of the form $x^n \ln x$ will always yield $n x^{n-1} \ln x + x^{n-1}$. This pattern makes reverse-engineering the integral extremely fast.

Answer 1

The Correct Option is B

Concept:
This problem can be solved using Integration by Parts ($\int u , dv = uv - \int v , du$). However, inspecting the multiple-choice options reveals a common structure ($x^n \log x$), suggesting that reverse differentiation (Product Rule) will be much faster. Let's use the differentiation method.
Step 1: Test Option B using the Product Rule.
The Product Rule is $\frac{d}{dx}(uv) = u'v + uv'$. Let's differentiate Option B: $y = x^2 \log x$. Let $u = x^2$ and $v = \log x$ (assuming natural log as is standard in calculus unless base 10 is specified). $$u' = 2x$$ $$v' = \frac{1}{x}$$
Step 2: Apply the formula.
$$y' = (2x)(\log x) + (x^2)\left(\frac{1}{x}\right)$$ $$y' = 2x \log x + x$$
Step 3: Factor the derivative to match the integrand.
Factor out an $x$ from both terms: $$y' = x(2\log x + 1)$$ This perfectly matches the given integrand $x(1 + 2\log x)$.
Step 4: Confirm the integral.
Since the derivative of $x^2\log x$ is the integrand, the integral of the integrand must be $x^2\log x + C$.