(a) Evaluate: ∫ { 𝟏 𝒍𝒐𝒈 𝒙 − 𝟏 (𝒍𝒐𝒈 𝒙)𝟐} 𝒅𝒙; (where𝒙 > 𝟏). OR

> [!TLDR] > The key idea is to rewrite the integrand as a derivative of a quotient using the product rule in reverse. The integral evaluates to $\frac{x}{\log x} + C$. Let’s understand why this integral is special. You have two terms: $\frac{1}{\log x}$ and $-\frac{1}{(\log x)^2}$. The presence of $\log x$ in the denominator suggests that differentiation of something like $\frac{x}{\log x}$ might produce these terms. Indeed, when you differentiate a quotient, you get two pieces — one from the numerator’s derivative and one from the denominator’s derivative. That’s exactly what we see here. So instead of hunting for a substitution, we can directly recognise that the integrand is the derivative of $\frac{x}{\log x}$. Let’s verify this step by step. 1. **Differentiate $\frac{x}{\log x}$ using the quotient rule.** Recall: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$. Here $u = x$, $v = \log x$. $u' = 1$, $v' = \frac{1}{x}$. So: \[ \frac{d}{dx}\left(\frac{x}{\log x}\right) = \frac{1 \cdot \log x - x \cdot \frac{1}{x}}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2}. \] 2. **Rewrite the derivative to match the integrand.** The expression $\frac{\log x - 1}{(\log x)^2}$ can be split: \[ \frac{\log x}{(\log x)^2} - \frac{1}{(\log x)^2} = \frac{1}{\log x} - \frac{1}{(\log x)^2}. \] That’s exactly the integrand we have: $\frac{1}{\log x} - \frac{1}{(\log x)^2}$. 3. **Therefore, the integral is simply the antiderivative we started with.** \[ \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx = \frac{x}{\log x} + C. \] > [!WARNING] > A common mistake is to try integration by parts or substitution unnecessarily. The integrand is already a perfect derivative — no extra work needed. Also, note the domain $x > 1$ ensures $\log x > 0$, so no division by zero issues. > [!TIP] > Whenever you see a sum/difference of terms with $\log x$ in denominators, check if the expression is the derivative of

(a) Evaluate: ∫ 𝟏 𝒍𝒐𝒈 𝒙 − 𝟏 (𝒍𝒐𝒈 𝒙)𝟐 𝒅𝒙; ( — CBSE Class 12 Mathematics 2025 PYQ

Question

Q. (a) Evaluate: \int { 𝟏 𝒍𝒐𝒈 𝒙 - 𝟏 (𝒍𝒐𝒈 𝒙)𝟐} 𝒅𝒙; (where𝒙 > 𝟏). OR

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✓ Free question

Concept understanding — Power Rule Integration

The Power Rule for Integration

Integration reverses differentiation: given a rate of change, it recovers the original function. When you differentiate $x^n$ you get $nx^{n-1}$ — the exponent drops by one and multiplies in front. To integrate you do the opposite: raise the exponent by one and divide by the new exponent. That is the whole idea.

The statement

\int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\qquad n\neq -1

$n$ may be any real number except $-1$ (fractions, negatives and $0$ all work).
$C$ is the constant of integration — shifting a graph up or down does not change its slope, so infinitely many functions share the same derivative.

Why it works

Differentiate the answer and you should get back the integrand:

\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}+C\right)=\frac{(n+1)x^{n}}{n+1}=x^{n}.

That one line is the proof.

Using it

\int x^3\,dx=\frac{x^4}{4}+C,\qquad \int \sqrt{x}\,dx=\int x^{1/2}\,dx=\frac{x^{3/2}}{3/2}+C=\frac{2}{3}x^{3/2}+C.

For a polynomial, apply it term by term:

\int(5x^3-2x+7)\,dx=\frac{5}{4}x^4-x^2+7x+C.

The one exception: $n=-1$

The formula needs $n+1\neq 0$ . For $n=-1$ it would divide by zero, so a different result takes over:

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

The absolute value keeps the logarithm defined for negative $x$ as well.

Watch out

The commonest slip is forgetting to divide by the new exponent — writing $\int x^3\,dx=x^4+C$ . Check by differentiating: $\dfrac{d}{dx}x^4=4x^3$ , not $x^3$ , so you must divide by $4$ .

The power rule for integration is the very first formula taught in the NCERT Class 12 Integrals chapter and underlies nearly every subsequent integration technique in CBSE boards and JEE Main. Students searching 'power rule of integration class 12 formula' or 'integration of xn examples' will find this raise-the-exponent-and-divide method, along with its log|x| exception at n = -1, is exactly what board exams test first.

The key idea is to rewrite the integrand as a derivative of a quotient using the product rule in reverse. The integral evaluates to $\frac{x}{\log x} + C$ .

Let’s understand why this integral is special. You have two terms: $\frac{1}{\log x}$ and $-\frac{1}{(\log x)^2}$ . The presence of $\log x$ in the denominator suggests that differentiation of something like $\frac{x}{\log x}$ might produce these terms. Indeed, when you differentiate a quotient, you get two pieces — one from the numerator’s derivative and one from the denominator’s derivative. That’s exactly what we see here.

So instead of hunting for a substitution, we can directly recognise that the integrand is the derivative of $\frac{x}{\log x}$ . Let’s verify this step by step.

Differentiate $\frac{x}{\log x}$ using the quotient rule. Recall: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ . Here $u = x$ , $v = \log x$ . $u' = 1$ , $v' = \frac{1}{x}$ . So:

\frac{d}{dx}\left(\frac{x}{\log x}\right) = \frac{1 \cdot \log x - x \cdot \frac{1}{x}}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2}.

Rewrite the derivative to match the integrand. The expression $\frac{\log x - 1}{(\log x)^2}$ can be split:

\frac{\log x}{(\log x)^2} - \frac{1}{(\log x)^2} = \frac{1}{\log x} - \frac{1}{(\log x)^2}.

That’s exactly the integrand we have: $\frac{1}{\log x} - \frac{1}{(\log x)^2}$ .

Therefore, the integral is simply the antiderivative we started with.

\int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx = \frac{x}{\log x} + C.

Watch out

A common mistake is to try integration by parts or substitution unnecessarily. The integrand is already a perfect derivative — no extra work needed. Also, note the domain $x > 1$ ensures $\log x > 0$ , so no division by zero issues.

Tip

Whenever you see a sum/difference of terms with $\log x$ in denominators, check if the expression is the derivative of $\frac{x}{\log x}$ or a similar rational function. This trick saves time in exams.

✓Final answer

\boxed{\frac{x}{\log x} + C}

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 41 on this concept.

CBSE 2026Set A1 markMCQ
$\int_1^4\frac{dx}{\sqrt{x}} =$
(a) $1$
(b) $-2$
(c) $2$
(d) $-1$
›Reveal solutionSolution

$\int_1^4\tfrac{dx}{\sqrt{x}}=[2\sqrt{x}]_1^4=4-2=2$ .

Since $\int x^{-1/2}\,dx=2\sqrt{x}$ , evaluate between the limits:

$[2\sqrt{x}]_1^4=2\sqrt{4}-2\sqrt{1}=2\cdot2-2\cdot1=4-2=2$ .

✓Final answer
$2$
CBSE 2026Set ANNUAL1 mark
$3x^2+4x^3$
›Reveal solutionSolution

Apply the power rule for integration to each term separately.

$\int (3x^2+4x^3)\,dx = 3\int x^2\,dx + 4\int x^3\,dx$

$= 3\cdot\frac{x^3}{3} + 4\cdot\frac{x^4}{4} + C$

$= x^3 + x^4 + C$

Check by differentiating: $\dfrac{d}{dx}(x^3+x^4+C) = 3x^2+4x^3$ ✓, matching the integrand.

✓Final answer
$x^3 + x^4 + C$
CBSE 2026Set ANNUAL1 mark
$\int_1^3 dx$
›Reveal solutionSolution

$\displaystyle\int_1^3 dx=[x]_1^3=3-1=2$ .

$\displaystyle\int_1^3 dx=[x]_1^3=3-1=2.$

✓Final answer
$2$
CBSE 2025Set 65/1/11 markMCQ
$\int \frac{2^{1/x}}{x^2} dx = k \cdot 2^{1/x} + C$
›Reveal solutionSolution
This problem asks for the coefficient $k$ in a given integral expression. We solve it by applying integration by parts to the left-hand side and comparing the resulting elementary term with the given form. The value of $k$ is $\boxed{-1}$ .

The problem asks us to find the value of $k$ given the equation $\int \frac{2^x}{x^2} dx = k \cdot 2^x \frac{1}{x} + C$ . The integral $\int \frac{2^x}{x^2} dx$ is a non-elementary integral, meaning it cannot be expressed in terms of elementary functions (polynomials, exponentials, logarithms, trigonometric functions).

In such problems, the given form $k \cdot 2^x \frac{1}{x} + C$ usually represents the elementary part obtained from a single application of integration by parts, with the remaining non-elementary integral implicitly absorbed or ignored for the purpose of finding $k$ . Our strategy will be to perform integration by parts on the left-hand side and then compare the resulting elementary term with $k \cdot 2^x \frac{1}{x}$ .

Recall the Integration by Parts Formula:

The integration by parts formula is given by:
$\int u \, dv = uv - \int v \, du$

Choose $u$ and $dv$ for the integral $\int \frac{2^x}{x^2} dx$ :

We have the integrand $\frac{2^x}{x^2} = 2^x \cdot x^{-2}$ .

To obtain a term like $2^x \frac{1}{x}$ in the $uv$ part, we should choose $dv$ such that $v$ involves $\frac{1}{x}$ .

Let's choose:

$dv = x^{-2} dx$

$u = 2^x$

Now, we find $v$ and $du$ :

$v = \int x^{-2} dx = -x^{-1} = -\frac{1}{x}$

$du = \frac{d}{dx}(2^x) dx = 2^x \log 2 \, dx$ (Recall that $\frac{d}{dx}(a^x) = a^x \log a$ )

Apply Integration by Parts:

Substitute these into the formula $\int u \, dv = uv - \int v \, du$ :

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

Compare with the given form: We are given that $\int \frac{2^x}{x^2} dx = k \cdot 2^x \frac{1}{x} + C$ . From our integration by parts, we found:

$\int \frac{2^x}{x^2} dx = -\frac{2^x}{x} + \log 2 \int \frac{2^x}{x} dx$
Comparing the elementary term involving $2^x \frac{1}{x}$: The term from our calculation is $-\frac{2^x}{x}$. The term given in the problem is $k \cdot 2^x \frac{1}{x}$. Therefore, we can equate these terms:
$k \cdot 2^x \frac{1}{x} = -\frac{2^x}{x}$
Dividing both sides by $\frac{2^x}{x}$ (which is non-zero for $x \neq 0$), we get:
$k = -1$
> [!WARNING] > A common mistake is to assume that the entire integral is exactly equal to $k \cdot 2^x \frac{1}{x} + C$. If this were true, differentiating $k \cdot 2^x \frac{1}{x} + C$ would yield $\frac{2^x}{x^2}$. However, $\frac{d}{dx} \left( k \cdot 2^x \frac{1}{x} \right) = k \cdot 2^x \left( \frac{\log 2}{x} - \frac{1}{x^2} \right)$. Equating this to $\frac{2^x}{x^2}$ leads to a contradiction ($1+k = kx \log 2$, which cannot hold for all $x$). This confirms that the problem implicitly asks for the coefficient of the elementary part obtained from the first step of integration by parts, with the remaining non-elementary integral (here, $\log 2 \int \frac{2^x}{x} dx$) not being part of the $k \cdot 2^x \frac{1}{x}$ term.

The value of $k$ is $-1$ .

✓Final answer
$k$
CBSE 2025Set X11 mark
$\int_{7}^{13} 1\,dx =$
›Reveal solutionSolution

$\displaystyle\int_{7}^{13} 1\,dx$ equals the interval length $13-7 = 6$ .

Concept: The definite integral of the constant function $1$ over $[a,b]$ equals $b-a$ .

Step 1 — Antiderivative.
$\int 1\,dx = x.$
Step 2 — Apply limits.
$\int_{7}^{13} 1\,dx = [x]_{7}^{13} = 13 - 7 = 6.$
✓Final answer
$6$
CBSE 2025Set ANNUAL1 mark
$\displaystyle\int \dfrac{1-x^3}{x^2}\,dx$
›Reveal solutionSolution

Split the integrand into two power-of- $x$ terms and integrate termwise.

$\int \dfrac{1-x^3}{x^2}\,dx = \int\left(x^{-2}-x\right)dx = -x^{-1}-\dfrac{x^2}{2}+c = -\dfrac1x-\dfrac{x^2}{2}+c$

✓Final answer
$-\dfrac1x - \dfrac{x^2}{2} + c$
CBSE 2025Set E1 markMCQ
$\int_{0}^{1} 3x^2\,dx =$
(a) $3$
(b) $\frac{1}{3}$
(c) $1$
(d) $\frac{1}{9}$
›Reveal solutionSolution

Integrate the power and apply the limits; the value is $1$ .

$\int_{0}^{1}3x^2\,dx=3\cdot\frac{x^3}{3}\Big|_0^1=[x^3]_0^1=1^3-0^3=1.$

✓Final answer
$1$
CBSE 2025Set E1 markMCQ
$\int x^m \cdot x^n\,dx =$
(a) $\frac{x^{m+1} \cdot x^{n+1}}{m + n + 2} + k$
(b) $\frac{x^{m+n}}{m + n} + k$
(c) $\frac{x^{m+n+1}}{m + n + 1} + k$
(d) $(m + n)x^{m+n-1} + k$
›Reveal solutionSolution

$x^m\cdot x^n = x^{m+n}$ , and $\int x^{m+n}\,dx = \frac{x^{m+n+1}}{m+n+1}+k$ .

Add the exponents:

$x^m\cdot x^n = x^{m+n}.$

Using the power rule $\int x^p\,dx = \frac{x^{p+1}}{p+1}+k$ with $p = m+n$ :

$\int x^{m+n}\,dx = \frac{x^{m+n+1}}{m+n+1} + k.$

✓Final answer
$\frac{x^{m+n+1}}{m+n+1} + k$
CBSE 2025Set E1 markMCQ
$\int \frac{(\sqrt{x} + 1)^2}{x\sqrt{x} + 2x + \sqrt{x}}\,dx =$
(a) $\sqrt{x} + k$
(b) $\frac{1}{2}\sqrt{x} + k$
(c) $2\sqrt{x} + k$
(d) $2x + k$
›Reveal solutionSolution

The integrand simplifies to $\frac{1}{\sqrt{x}}$ , whose integral is $2\sqrt{x}+k$ .

Factor the denominator:

$x\sqrt{x} + 2x + \sqrt{x} = \sqrt{x}\left(x + 2\sqrt{x} + 1\right) = \sqrt{x}\,(\sqrt{x} + 1)^2.$

So the integrand is

$\frac{(\sqrt{x}+1)^2}{\sqrt{x}\,(\sqrt{x}+1)^2} = \frac{1}{\sqrt{x}}.$

Therefore

$\int \frac{dx}{\sqrt{x}} = 2\sqrt{x} + k.$

✓Final answer
$2\sqrt{x} + k$
CBSE 2025Set E1 markMCQ
$\int_{0}^{a} \frac{dx}{\sqrt{x}} =$
(a) $2\sqrt{x}$
(b) $2\sqrt{a}$
(c) $\sqrt{x}$
(d) $\sqrt{a}$
›Reveal solutionSolution

$\int_0^a x^{-1/2}\,dx = [2\sqrt{x}]_0^a = 2\sqrt{a}$ .

Use the power rule $\int x^{-1/2}dx = 2x^{1/2}=2\sqrt{x}$ . Evaluating the definite integral:

$\int_0^a \frac{dx}{\sqrt{x}} = \big[2\sqrt{x}\big]_0^a = 2\sqrt{a} - 0 = 2\sqrt{a}.$

✓Final answer
$2\sqrt{a}$
CBSE 2025Set E1 markMCQ
$2\int_{1}^{9} \frac{dx}{\sqrt{x}} =$
(a) $8$
(b) $4$
(c) $2$
(d) $12$
›Reveal solutionSolution

$2[2\sqrt{x}]_1^9 = 2(2\cdot3 - 2\cdot1) = 2\cdot4 = 8$ .

Use $\int x^{-1/2}dx = 2\sqrt{x}$ :

$2\int_1^9 \frac{dx}{\sqrt{x}} = 2\big[2\sqrt{x}\big]_1^9 = 2\big(2\sqrt{9}-2\sqrt{1}\big) = 2(6-2) = 8.$

✓Final answer
$8$
CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\int x^{5/3}\, dx =$
(a) $\dfrac{3}{5}x^{2/3} + c$
(b) $\dfrac{8}{3}x^{8/3} + c$
(c) $\dfrac{3}{8}x^{8/3} + c$
(d) $\dfrac{5}{3}x^{8/3} + c$
›Reveal solutionSolution

Apply the standard power rule for integration, $\int x^n dx = \dfrac{x^{n+1}}{n+1}+c$ .

$\displaystyle\int x^{5/3}\,dx = \dfrac{x^{5/3+1}}{5/3+1}+c = \dfrac{x^{8/3}}{8/3}+c = \dfrac{3}{8}x^{8/3}+c$

✓Final answer
$\dfrac{3}{8}x^{8/3}+c$

Q.(a) Evaluate: ∫ { 𝟏 𝒍𝒐𝒈 𝒙 − 𝟏 (𝒍𝒐𝒈 𝒙)𝟐} 𝒅𝒙; (where𝒙 > 𝟏). OR

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