Q.Evaluate the integral using substitution $\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2x^2}\right)e^{2x}\,dx$

The key idea is to notice that the derivative of $e^{2x}$ is $2e^{2x}$, but here the factor in front is $\frac{1}{x} - \frac{1}{2x^2}$. This suggests checking if the expression is the derivative of $\frac{e^{2x}}{2x}$.

Differentiate $\frac{e^{2x}}{2x}$ using the quotient rule:

$\frac{d}{dx}\left(\frac{e^{2x}}{2x}\right) = \frac{2e^{2x} \cdot 2x - e^{2x} \cdot 2}{(2x)^2} = \frac{4x e^{2x} - 2e^{2x}}{4x^2} = \frac{2x e^{2x} - e^{2x}}{2x^2} = e^{2x}\left(\frac{1}{x} - \frac{1}{2x^2}\right).$ …

We evaluate $\displaystyle\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2x^2}\right)e^{2x}\,dx$.

1. Recognise the exact derivative. For the function $\dfrac{e^{2x}}{2x}$,

$\frac{d}{dx}\!\left(\frac{e^{2x}}{2x}\right)=\frac{(2x)(2e^{2x})-e^{2x}(2)}{4x^2}=\frac{e^{2x}(2x-1)}{2x^2}=\left(\frac{1}{x}-\frac{1}{2x^2}\right)e^{2x}.$ …

Method: Recognise an exact derivative — the $e^{ax}\{f(x)+f'(x)\}$ pattern

Some integrands are already the derivative of a product; spotting this skips all technique. The classic template is $\displaystyle\int e^{ax}\big(f(x)+\tfrac{1}{a}f'(x)\big)\,dx$-type combinations that reassemble into a single product.

Steps

Step 1: Suspect an exact derivative.

When an exponential multiplies a function and something resembling its derivative, guess an antiderivative of the form $\dfrac{e^{ax}}{\text{something}}$ or $e^{ax}f(x)$.

Step 2: Verify by differentiating your guess. …

Common Mistakes

Mistake 1: Attempting a plain $u$-substitution or by-parts and getting stuck in a loop.

Why it's wrong: the integrand is already an exact derivative of $\tfrac{e^{2x}}{2x}$, so brute-force methods spiral; recognising the pattern $e^{ax}\{f+f'\}$ is the intended route. Correct approach: verify $\frac{d}{dx}\!\left(\tfrac{e^{2x}}{2x}\right)$ equals the integrand.

Mistake 2: Slipping on the quotient-rule differentiation when checking. …