Evaluate $\int_{2}^{3} \dfrac{2x}{1 + x^2}\, dx$.

Evaluate ∫₂³ (2x)/(1 + x²) dx.

Question of 373

\int_{2}^{3} \dfrac{2x}{1 + x^2}\, dx

Telangana TsbieTelangana Board of Intermediate Education 2023Subjective· 2mImportance★★★★★

0% · 0/373 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Natural Logarithm Integration

Natural Logarithm Integration: From Intuition to Formula

You already know that integration is the reverse of differentiation. So the first question is: what function, when differentiated, gives $\frac{1}{x}$ ?

You know $\frac{d}{dx}(x^n) = n x^{n-1}$ . Trying to find a function whose derivative is $x^{-1}$ , the power rule would give $\frac{x^0}{0}$ , which is undefined. That's the clue — $\frac{1}{x}$ doesn't fit the power rule pattern.

The function that fills this gap is the natural logarithm, $\log x$ . Its derivative is exactly $\frac{1}{x}$ (for $x > 0$ ), so integration reverses this:

\int \frac{1}{x} \, dx = \log|x| + C

The absolute value $|x|$ is crucial — it extends the formula to negative $x$ , because $\log x$ is only defined for positive numbers, but $\frac{1}{x}$ is defined for all $x \neq 0$ .

Why the absolute value?

For $x > 0$ , differentiating $\log|x|$ gives $\frac{1}{x}$ . For $x < 0$ , $\log|x| = \log(-x)$ , and its derivative is $\frac{1}{-x} \cdot (-1) = \frac{1}{x}$ . Same result, so $\log|x|$ works for both sides.

The generalised form

The real power comes when the numerator is the derivative of the denominator:

\int \frac{f'(x)}{f(x)} \, dx = \log|f(x)| + C

This is the logarithmic integration pattern.

Example to see it in action

Find $\int \frac{2x}{x^2 + 1} \, dx$ . Here $f(x) = x^2 + 1$ , so $f'(x) = 2x$ — the numerator matches. Therefore:

\int \frac{2x}{x^2 + 1} \, dx = \log|x^2 + 1| + C = \log(x^2 + 1) + C

(dropping the absolute value since $x^2 + 1$ is always positive).

What if the numerator doesn't match exactly?

For $\int \frac{x}{x^2 + 1} \, dx$ , the derivative of the denominator is $2x$ but you only have $x$ . Adjust by factoring:

\int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \int \frac{2x}{x^2 + 1} \, dx = \frac{1}{2} \log|x^2 + 1| + C

Tip

When the numerator is a constant multiple of the derivative of the denominator, factor out that constant: $\int \frac{k \cdot f'(x)}{f(x)} \, dx = k \log|f(x)| + C$ .

Common mistake to avoid …

Start your 14-day free trial to unlock the full solution →

Since $\tfrac{d}{dx}(1+x^2)=2x$ , the integral is $\ln(1+x^2)$ ; evaluating gives $\ln 2$ .

The numerator is exactly the derivative of the denominator, so

$\int_2^3 \dfrac{2x}{1+x^2}\,dx = \big[\ln(1 + x^2)\big]_2^3$ .

…

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 25 on this concept.

CBSE 2026Set A1 markMCQ
$\int\frac{dx}{x - 1} =$
(a) $\log|x + 1| + k$
(b) $-\log|x + 1| + k$
(c) $\log|x - 1| + k$
(d) $\log x + k$
›Reveal solutionSolution

Standard log integral: $\int\tfrac{dx}{x-1}=\log|x-1|+k$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. \int tan x dx equals$
(a) log|sin x| + C
(b) log|cos x| + C
(c) log|sec x| + C
(d) log|cot x| + C
›Reveal solutionSolution

Write $\tan x$ as $\sin x/\cos x$ and integrate by substitution $u=\cos x$ .

$\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx$

Let $u=\cos x$ , so $du=-\sin x\,dx$ , i.e. $\sin x\,dx = -du$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$\int \frac{dx}{x\log x} =$
(a) $\log x + k$
(b) $(\log x)^2 + k$
(c) $\log(\log x) + k$
(d) $\frac{1}{\log x} + k$
›Reveal solutionSolution

With $t = \log x$ , the integral becomes $\int \frac{dt}{t} = \log(\log x)+k$ .

Let $t = \log x$ , so $dt = \frac{1}{x}\,dx$ . Then …

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$\int \frac{x - 3}{x^2 - 9}\,dx =$
(a) $\log(x - 3) + k$
(b) $\log(x + 3) + k$
(c) $-\frac{1}{(x + 3)^2} + k$
(d) $\frac{x^2}{2} - 3x + k$
›Reveal solutionSolution

$\frac{x-3}{x^2-9} = \frac{1}{x+3}$ , so the integral is $\log(x+3)+k$ .

Factor the denominator using $x^2 - 9 = (x-3)(x+3)$ :

$\frac{x-3}{x^2-9} = \frac{x-3}{(x-3)(x+3)} = \frac{1}{x+3}.$ …

Start your 14-day free trial to see the full solution →
CBSE 2024Set D1 markMCQ
$\int \sec x\,dx =$
(a) $\log|\sec x + \tan x| + c$
(b) $\log|\sec x - \tan x| + c$
(c) $\log \sec x + c$
(d) $\tan^5 x + c$
›Reveal solutionSolution

$\int\sec x\,dx=\log|\sec x+\tan x|+c$ .

Multiplying numerator and denominator by $(\sec x+\tan x)$ and substituting $t=\sec x+\tan x$ gives the standard result:

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set D1 markMCQ
$\int \frac{dx}{2 - 3x} =$
(a) $-3\log|2 - 3x| + c$
(b) $-\frac{1}{3}\log|2 - 3x| + c$
(c) $-\log|2 - 3x| + c$
(d) $2\tan^{-1} x^4 + c$
›Reveal solutionSolution

$\int \frac{dx}{a+bx} = \frac{1}{b}\log|a+bx| + c$ ; here $b = -3$ .

Let $u = 2 - 3x$ , so $du = -3\,dx$ , i.e. $dx = -\frac{1}{3}du$ . Then …

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$\displaystyle\int \dfrac{\sin x}{1 + \cos x}\,dx$
›Reveal solutionSolution

Substitute $u = 1+\cos x$ ; then $du = -\sin x\,dx$ , turning the integral into a simple $\dfrac{du}{u}$ form.

$\displaystyle\int \dfrac{\sin x}{1+\cos x}\, dx$

Let $u = 1 + \cos x$ , so $du = -\sin x\, dx$ , i.e. $\sin x\, dx = -du$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$\int_2^3 \frac{1}{x} \,dx =$
(a) $\frac{2}{3}\log x$
(b) $\log\left(\frac{3}{2}\right)$
(c) $\log\left(\frac{2}{3}\right)$
(d) none of these
›Reveal solutionSolution

Integrate 1/x to get log x, then apply the limits and combine using log rules.

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$\displaystyle\int \dfrac{dx}{3x+2}$
›Reveal solutionSolution

Direct substitution $u=3x+2$ .

$\int\dfrac{dx}{3x+2}$

Let $u=3x+2 \implies du=3\,dx \implies dx=\dfrac{du}{3}$ . …

Start your 14-day free trial to see the full solution →
CBSE 2023Set E1 markMCQ
$\displaystyle\int\dfrac{x+2}{x^{2}-4}\,dx =$
(a) $\log|x+2|+k$
(b) $\log|x^{2}-4|+k$
(c) $\log|x-2|+k$
(d) $\log\left|\dfrac{x+2}{x-2}\right|+k$
›Reveal solutionSolution

Cancelling $(x+2)$ reduces the integrand to $\dfrac{1}{x-2}$ , whose integral is $\log|x-2|+k$ .

Factor the denominator: $x^{2}-4=(x-2)(x+2)$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set E1 markMCQ
$\displaystyle\int\dfrac{3\cos x-2\sin x}{2\cos x+3\sin x}\,dx =$
(a) $2\cos x+3\sin x+k$
(b) $\log|2\cos x+3\sin x|+k$
(c) $\tan^{-1}\left(3\sin\dfrac{x}{2}\right)+k$
(d) $2\tan\dfrac{x}{2}+k$
›Reveal solutionSolution

Since $\dfrac{d}{dx}(2\cos x+3\sin x)=3\cos x-2\sin x$ , the integral is of the form $\int\tfrac{f'}{f}=\log|f|$ .

Let $f(x)=2\cos x+3\sin x$ . Then $f'(x)=-2\sin x+3\cos x=3\cos x-2\sin x$ , which is the numerator.

Using $\int\dfrac{f'(x)}{f(x)}\,dx=\log|f(x)|+k$ : …

Start your 14-day free trial to see the full solution →
CBSE 2023Set E1 markMCQ
$\displaystyle\int\dfrac{3x^{2}+2}{x^{3}+2x}\,dx =$
(a) $\sin^{-1}(x^{3}+2x)+k$
(b) $\tan^{-1}(3x^{2}+2)+k$
(c) $\log|3x^{2}+2|+k$
(d) $\log|x^{3}+2x|+k$
›Reveal solutionSolution

Since $\dfrac{d}{dx}(x^{3}+2x)=3x^{2}+2$ , the integral is $\int\tfrac{f'}{f}=\log|f|$ .

Let $f(x)=x^{3}+2x$ . Then $f'(x)=3x^{2}+2$ , exactly the numerator.

Using $\int\dfrac{f'(x)}{f(x)}\,dx=\log|f(x)|+k$ : …

Start your 14-day free trial to see the full solution →

Q.Evaluate ∫232x1+x2 dx\int_{2}^{3} \dfrac{2x}{1 + x^2}\, dx∫23​1+x22x​dx.

🎓Unlock everything free for 14 days

Q.Evaluate $\int_{2}^{3} \dfrac{2x}{1 + x^2}\, dx$ .

Unlock everything free for 14 days