Sample 2026-27 Class 12 Maths Q12 Solution: Find P in ∫ 1/(1 + e²ˣ) dx = x − P log(1 + e²ˣ) + C

The question

If $\int \frac{1}{1 + e^{2x}}\,dx = x - P\log(1 + e^{2x}) + C$ ⁠, then ‘ $P$ ’ is equal to

(A) $-\frac{1}{2}$

(B) $-\frac{1}{3}$

(C) $\frac{1}{3}$

(D) $\frac{1}{2}$

Exam answer

\begin{aligned} &\int \frac{1}{1 + e^{2x}}\,dx \\ &= \int \frac{(1 + e^{2x}) - e^{2x}}{1 + e^{2x}}\,dx \\ &= \int \left(1 - \frac{e^{2x}}{1 + e^{2x}}\right)dx \\ &= x - \int \frac{e^{2x}}{1 + e^{2x}}\,dx \end{aligned}

Why?

As it stands, $\frac{1}{1 + e^{2x}}$ has no standard integral. Writing the numerator 1 as $(1 + e^{2x}) - e^{2x}$ splits it into 1, which is easy, and a fraction whose numerator is almost the derivative of its denominator.

Split the fraction into two, over the same denominator:

\begin{aligned} &\frac{(1 + e^{2x}) - e^{2x}}{1 + e^{2x}} \\ &= \frac{1 + e^{2x}}{1 + e^{2x}} - \frac{e^{2x}}{1 + e^{2x}} \\ &= 1 - \frac{e^{2x}}{1 + e^{2x}} \end{aligned}

Then $\int 1\,dx = x$ ⁠.

Putting $1 + e^{2x} = t$ ⁠, $2e^{2x}\,dx = dt$ ⁠:

\begin{aligned} \int \frac{e^{2x}}{1 + e^{2x}}\,dx &= \frac{1}{2}\int \frac{dt}{t} \\ &= \frac{1}{2}\log|t| \\ &= \frac{1}{2}\log|1 + e^{2x}| \end{aligned}

Why?

The derivative of $1 + e^{2x}$ is $2e^{2x}$ ⁠, so $e^{2x}\,dx = \frac{1}{2}\,dt$ ⁠. Then use the standard integral

\begin{aligned} \frac{d}{dx}(1 + e^{2x}) &= 0 + e^{2x} \cdot 2 \\ dt &= 2e^{2x}\,dx \\ \frac{1}{2}\,dt &= e^{2x}\,dx \\[4pt] \int \frac{e^{2x}\,dx}{1 + e^{2x}} &= \int \frac{\frac12\,dt}{t} \\ &= \frac{1}{2}\int \frac{dt}{t} \end{aligned}

\int \frac{1}{t}\,dt = \log|t| + C

\begin{aligned} &\int \frac{1}{1 + e^{2x}}\,dx \\ &= x - \frac{1}{2}\log|1 + e^{2x}| + C \end{aligned}

Comparing with $x - P\log(1 + e^{2x}) + C$ (here $1 + e^{2x} > 0$ ⁠), $P = \frac{1}{2}$ ⁠.

Why?

$1 + e^{2x}$ is always positive, so $|1 + e^{2x}| = 1 + e^{2x}$ and the two forms match term by term.

Watch out: the minus sign is already written in the question, so $P = +\frac{1}{2}$ ⁠, not $-\frac{1}{2}$ as in (A).

Answer: (D) $\dfrac{1}{2}$

Where this comes from

Chapter 7, Integrals: Integration by substitution.

Question type: Substitution

How to spot it: click to see

Type 2. Substitution: when the derivative is hiding in the question

Spot it: a function and its derivative both appear, like $\log x$ and $\frac{1}{x}$ ⁠, or $x^2 + 1$ and $x$ ⁠.

Here: Writing the 1 on top as $(1 + e^{2x}) - e^{2x}$ leaves $x$ minus $\int\frac{e^{2x}}{1 + e^{2x}}\,dx$ ⁠, where the top is half the derivative of the bottom. So put $1 + e^{2x} = t$ ⁠, which gives $\frac12\log(1 + e^{2x})$ and $P = \frac12$ ⁠.

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Q12: Find PPP in ∫11+e2x dx=x−Plog⁡(1+e2x)+C\int \frac{1}{1 + e^{2x}}\,dx = x - P\log(1 + e^{2x}) + C∫1+e2x1​dx=x−Plog(1+e2x)+C

Q12: Find $P$ in $\int \frac{1}{1 + e^{2x}}\,dx = x - P\log(1 + e^{2x}) + C$