Question

$\int \frac{3^x(x\log 3 - 1)}{x^2} dx =$

• A. $x \cdot 3^x + c$
• B. $\frac{3^x}{x^2} + c$
• C. $x^2 3^x + c$
• D. $\frac{3^x}{x} + c$

Hint:

When an integrand looks complicated, especially if it's a fraction with a squared denominator, check if it matches the result of a standard differentiation rule like the quotient rule or product rule. Recognizing these patterns can turn a difficult integration into a simple reverse differentiation.

Answer 1

The Correct Option is D

The integrand $\frac{3^x(x\log 3 - 1)}{x^2}$ has the form of the result of the quotient rule for differentiation.
Let's consider the function $f(x) = \frac{u(x)}{v(x)}$ where $u(x)=3^x$ and $v(x)=x$.
According to the quotient rule, the derivative is $f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$.
First, find the derivatives of $u(x)$ and $v(x)$:
$u'(x) = \frac{d}{dx}(3^x) = 3^x \log 3$.
$v'(x) = \frac{d}{dx}(x) = 1$.
Now, apply the quotient rule:
$\frac{d}{dx}\left(\frac{3^x}{x}\right) = \frac{x(3^x \log 3) - 3^x(1)}{x^2}$.
$\frac{d}{dx}\left(\frac{3^x}{x}\right) = \frac{3^x(x \log 3 - 1)}{x^2}$.
This is exactly the integrand given in the question.
Therefore, the integral of this expression is the original function from which it was derived.
$\int \frac{3^x(x\log 3 - 1)}{x^2} dx = \frac{3^x}{x} + C$.