Q.Integrate the function $\frac{e^x}{(1+e^x)(2+e^x)}$

The key idea is Partial Fraction Decomposition applied to a function of $e^x$, followed by a simple substitution.

Let $t = e^x$, so $dt = e^x dx$. The integral becomes

Decompose $\frac{1}{(1+t)(2+t)}$ into partial fractions:

Integrate term by term: …

Why Partial Fractions Work Here

When you see a product of linear factors in the denominator — like $(1+e^x)(2+e^x)$ — and a numerator that is essentially the derivative of one of those factors, your first instinct should be substitution. Here, $e^x$ is both the numerator and the derivative of $e^x$ itself. That’s a strong hint: let $t = e^x$, so $dt = e^x dx$, and the integral becomes a clean rational function in $t$.

The denominator becomes $(1+t)(2+t)$, and the numerator is just $dt$. So we’re integrating $\frac{1}{(1+t)(2+t)}\, dt$. This is a textbook partial fractions problem: split the fraction into two simpler pieces, each of which integrates to a logarithm.

Step-by-Step Solution

1. Substitute $t = e^x$

Let $t = e^x$. Then $dt = e^x dx$, which is exactly the numerator of our integrand. So:

2. Set up partial fractions

We want to write:

Multiply both sides by $(1+t)(2+t)$:

3. Solve for $A$ and $B$

We can solve by choosing convenient values of $t$:

• Let $t = -1$: then $1 = A(2-1) + B(0) \implies 1 = A \cdot 1 \implies A = 1$
• Let $t = -2$: then $1 = A(0) + B(1-2) \implies 1 = B \cdot (-1) \implies B = -1$

4. Rewrite the integral

Now we have:

5. Integrate term by term

Each term integrates to a natural logarithm: …

Method: Substitute for the exponential, then partial-fraction

Use this when $e^x$ (or $a^x$) appears both as the numerator and inside the denominator's factors: let $t=e^x$ to turn the integral into a rational function.

Steps

Step 1: Substitute $t=e^x$.

Then $dt=e^x\,dx$. The numerator $e^x\,dx$ becomes exactly $dt$, and the denominator becomes a polynomial in $t$.

Step 2: Decompose into partial fractions. …

Common Mistakes

Mistake 1: Substituting $t=e^x$ but forgetting $dt=e^x\,dx$.

Why it's wrong: the numerator's $e^x\,dx$ is exactly $dt$; without noticing, students leave an extra $e^x$ unaccounted for. Correct approach: match the numerator to $dt$.

Mistake 2: Sign error solving for $B$. …