Find $\int \log x\, dx$

> [!TLDR] > The integral of $\log x$ is solved using integration by parts, treating $\log x$ as $1 \cdot \log x$. The result is $x \log x - x + C$. The key insight here is that $\log x$ doesn't have an obvious antiderivative from the power rule or standard formulas. But we can rewrite it as a product: $\log x = 1 \cdot \log x$. This lets us use **integration by parts**, which is the reverse of the product rule for derivatives. Integration by parts says: $\int u \, dv = uv - \int v \, du$. The trick is choosing $u$ and $dv$ so that the new integral $\int v \, du$ is simpler than the original. For $\log x$, we set $u = \log x$ because its derivative is $\frac{1}{x}$, a simple rational function. Then $dv = 1 \, dx$, so $v = x$. Let's work through it step by step. 1. **Set up integration by parts.** We have $\int \log x \, dx = \int (1)(\log x) \, dx$. Choose: $u = \log x$ $dv = 1 \, dx$ 2. **Find $du$ and $v$.** Differentiate $u$: $du = \frac{1}{x} \, dx$ Integrate $dv$: $v = \int 1 \, dx = x$ 3. **Apply the integration by parts formula.** $\int u \, dv = uv - \int v \, du$ Substitute: $\int \log x \, dx = (\log x)(x) - \int x \cdot \frac{1}{x} \, dx$ 4. **Simplify the new integral.** $x \cdot \frac{1}{x} = 1$, so we get: $\int \log x \, dx = x \log x - \int 1 \, dx$ 5. **Integrate the remaining term.** $\int 1 \, dx = x + C$ (don't forget the constant of integration) Therefore: $\int \log x \, dx = x \log x - x + C$ > [!WARNING] > A common mistake is to forget the constant of integration $C$ or to misapply the formula by swapping $u$ and $dv$. If you set $u = 1$ and $dv = \log x \, dx$, you'd need to know the integral of $\log x$ already — which is exactly what we're trying to find! Always choose $u$ as the function that simplifies when differentiated. > [!TIP] > This result is a classic and worth memorizing: $\int \log x \, dx = x \log x - x + C$. It also works for $\ln x$ (natural log) —

Find ∫ log x dx

Worked Examples · Example 18

\int \log x\, dx

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✓ Free question

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

Do not apply this pattern when the numerator is unrelated to the derivative of the denominator. For example, $\int \frac{1}{x^2 + 1} \, dx$ is not $\log|x^2 + 1|$ — it gives $\tan^{-1} x + C$ , a completely different result.

Watch out

The rule only works when the numerator is exactly (or a constant multiple of) the derivative of the denominator. If not, use another method (partial fractions, trigonometric substitution, etc.).

Final formula to remember:

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

And the simplest case: $\int \frac{1}{x} \, dx = \log|x| + C$ .

The ∫f'(x)/f(x) dx = log|f(x)| + C pattern is one of the most tested standard results in the NCERT Class 12 Integrals chapter, appearing constantly in CBSE board and JEE Main 'evaluate the integral' questions. Students searching 'integration of 1/x formula' or 'logarithmic integration examples class 12' will find this numerator-matches-derivative-of-denominator rule is exactly the shortcut those exam papers expect students to spot instantly.

The integral of $\log x$ is solved using integration by parts, treating $\log x$ as $1 \cdot \log x$ . The result is $x \log x - x + C$ .

The key insight here is that $\log x$ doesn't have an obvious antiderivative from the power rule or standard formulas. But we can rewrite it as a product: $\log x = 1 \cdot \log x$ . This lets us use integration by parts, which is the reverse of the product rule for derivatives.

Integration by parts says: $\int u \, dv = uv - \int v \, du$ . The trick is choosing $u$ and $dv$ so that the new integral $\int v \, du$ is simpler than the original. For $\log x$ , we set $u = \log x$ because its derivative is $\frac{1}{x}$ , a simple rational function. Then $dv = 1 \, dx$ , so $v = x$ .

Let's work through it step by step.

Set up integration by parts.

We have $\int \log x \, dx = \int (1)(\log x) \, dx$ .

Choose:

$u = \log x$

$dv = 1 \, dx$
Find $du$ and $v$ .

Differentiate $u$ : $du = \frac{1}{x} \, dx$

Integrate $dv$ : $v = \int 1 \, dx = x$
Apply the integration by parts formula.

$\int u \, dv = uv - \int v \, du$

Substitute:

$\int \log x \, dx = (\log x)(x) - \int x \cdot \frac{1}{x} \, dx$
Simplify the new integral.

$x \cdot \frac{1}{x} = 1$ , so we get:

$\int \log x \, dx = x \log x - \int 1 \, dx$
Integrate the remaining term.

$\int 1 \, dx = x + C$ (don't forget the constant of integration)

Therefore:

$\int \log x \, dx = x \log x - x + C$

Watch out

A common mistake is to forget the constant of integration $C$ or to misapply the formula by swapping $u$ and $dv$ . If you set $u = 1$ and $dv = \log x \, dx$ , you'd need to know the integral of $\log x$ already — which is exactly what we're trying to find! Always choose $u$ as the function that simplifies when differentiated.

Tip

This result is a classic and worth memorizing: $\int \log x \, dx = x \log x - x + C$ . It also works for $\ln x$ (natural log) — same formula. For $\log_a x$ , use the change of base: $\log_a x = \frac{\ln x}{\ln a}$ , then integrate.

✓Final answer

\boxed{x \log x - 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 29 on this concept.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$f_n(x) = \displaystyle\int \dfrac{1-\log x}{x^n}\,dx$
›Reveal solutionSolution
This tests integration by parts on a family $f_n(x)=\int(1-\log x)x^{-n}dx$ and simplifying $f_2(e)-f_3(e)$ ; the answer is $\dfrac{4e-1}{4e^2}$ .

Concept and Intuition

The integrand $(1-\log x)x^{-n}$ is exactly the derivative of $\dfrac{\log x}{x^{n-1}}$ type expressions once you notice the pattern: differentiating $\dfrac{\log x}{x^{k}}$ produces a $\dfrac{1}{x^{k+1}}$ term and a $-k\dfrac{\log x}{x^{k+1}}$ term, which combine nicely for specific $n$ . Rather than guess, integrating by parts systematically gives a general formula valid for every $n\ne 1$ .

Step-by-Step Solution

Write $f_n(x)=\int (1-\log x)x^{-n}dx$ . Take $u=1-\log x$ , $dv=x^{-n}dx$ , so $du=-\dfrac1x dx$ , $v=\dfrac{x^{1-n}}{1-n}$ .

By parts: $f_n(x)=\dfrac{(1-\log x)x^{1-n}}{1-n}-\int \dfrac{x^{1-n}}{1-n}\left(-\dfrac1x\right)dx=\dfrac{(1-\log x)x^{1-n}}{1-n}+\dfrac{1}{1-n}\int x^{-n}dx$ .

$\int x^{-n}dx=\dfrac{x^{1-n}}{1-n}$ , so $f_n(x)=\dfrac{x^{1-n}}{1-n}\left[(1-\log x)+\dfrac{1}{1-n}\right]+C$ .

For $n=2$ ( $1-n=-1$ ): $f_2(x)=-x^{-1}\left[(1-\log x)-1\right]=-x^{-1}(-\log x)=\dfrac{\log x}{x}+C$ .

For $n=3$ ( $1-n=-2$ ): $f_3(x)=\dfrac{x^{-2}}{-2}\left[(1-\log x)-\dfrac12\right]=\dfrac{x^{-2}}{-2}\left(\dfrac12-\log x\right)=\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}+C$ .

Evaluate at $x=e$ : $f_2(e)=\dfrac{1}{e}$ , $f_3(e)=\dfrac{1}{2e^2}-\dfrac{1}{4e^2}=\dfrac{1}{4e^2}$ .

$f_2(e)-f_3(e)=\dfrac1e-\dfrac{1}{4e^2}=\dfrac{4e-1}{4e^2}=\dfrac{1}{4e^2}[4e-1]$ .

Common Mistakes

Forgetting the extra $\dfrac{1}{1-n}$ term generated from integrating $x^{-n}$ again during parts.

Sign errors when $1-n$ is negative (as it is for both $n=2,3$ ).

✓Final answer
$\dfrac{1}{4e^2}[4e-1]+c$

ANSWER: A
AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$\displaystyle\int \dfrac{e^{2x}-1}{e^{2x}+e^x+1}\,dx =$
›Reveal solutionSolution
This tests recognizing a disguised $\int \frac{f'(x)}{f(x)}dx$ form after dividing through by $e^x$ ; the answer is $\log(e^{2x}+e^x+1)-x+c$ .

Concept and Intuition

Whenever an integrand is a ratio of exponentials, dividing numerator and denominator by the smallest power often reveals a $\dfrac{du}{u}$ structure, since exponential derivatives reproduce exponentials. Here $e^{2x}+e^x+1$ divided by $e^x$ becomes $e^x+1+e^{-x}$ , and its derivative is $e^x-e^{-x}$ — exactly the (rescaled) numerator.

Step-by-Step Solution

Divide numerator and denominator of $\dfrac{e^{2x}-1}{e^{2x}+e^x+1}$ by $e^x$ :

$\frac{e^{2x}-1}{e^{2x}+e^x+1} = \frac{e^x - e^{-x}}{e^x+1+e^{-x}}$

Let $u = e^x + e^{-x} + 1$ . Then $\dfrac{du}{dx} = e^x - e^{-x}$ , which is exactly the numerator.

So the integral becomes $\displaystyle\int \frac{du}{u} = \log|u| + c = \log(e^x+e^{-x}+1) + c$ .

Rewrite in terms of $e^{2x}$ : multiply inside the log by $\dfrac{e^x}{e^x}$ :

$e^x+e^{-x}+1 = \frac{e^{2x}+1+e^x}{e^x}$

So $\log(e^x+e^{-x}+1) = \log(e^{2x}+e^x+1) - \log(e^x) = \log(e^{2x}+e^x+1) - x$ .

Final result: $\log(e^{2x}+e^x+1) - x + c$ .

Check by differentiating: $\dfrac{d}{dx}\big[\log(e^{2x}+e^x+1)-x\big] = \dfrac{2e^{2x}+e^x}{e^{2x}+e^x+1} - 1 = \dfrac{2e^{2x}+e^x-e^{2x}-e^x-1}{e^{2x}+e^x+1} = \dfrac{e^{2x}-1}{e^{2x}+e^x+1}$ , which matches the original integrand exactly.

Common Mistakes

Trying to integrate directly without dividing by $e^x$ first — the denominator's true derivative structure is hidden until this step.

Forgetting the $-x$ term that comes from converting $\log(e^x+e^{-x}+1)$ back into $\log(e^{2x}+e^x+1)$ form, which flips option (A) into the wrong sign.

✓Final answer
$\log(e^{2x}+e^x+1) - x + c$

ANSWER: B
AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$\displaystyle\int \dfrac{dx}{\cos x + \sqrt{3}\sin x}=$
›Reveal solutionSolution
Rewriting $\cos x+\sqrt3\sin x$ as $2\sin(x+\pi/6)$ turns the integral into the standard $\int\csc\theta\,d\theta$ form, giving $\frac12\log\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)+c$ .

Concept and Intuition

Any expression $a\cos x+b\sin x$ can be written as $R\sin(x+\phi)$ (or $R\cos(x-\phi)$ ) where $R=\sqrt{a^2+b^2}$ . This is the key move that converts a linear combination of sine and cosine into the standard $\int\csc\theta\,d\theta = \log|\tan(\theta/2)|+c$ form.

Step-by-Step Solution

$\cos x + \sqrt3\sin x = 2\left(\dfrac12\cos x + \dfrac{\sqrt3}{2}\sin x\right) = 2\left(\sin x\cos\dfrac{\pi}{6}+\cos x\sin\dfrac{\pi}{6}\right) = 2\sin\left(x+\dfrac{\pi}{6}\right)$ .

So $\displaystyle\int\dfrac{dx}{\cos x+\sqrt3\sin x} = \dfrac12\int\csc\left(x+\dfrac{\pi}{6}\right)dx$ .

Using $\displaystyle\int\csc\theta\,d\theta = \log\left|\tan\dfrac{\theta}{2}\right|+c$ with $\theta = x+\dfrac{\pi}{6}$ : $= \dfrac12\log\left|\tan\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)\right|+c$ .

Common Mistakes

Using $R\cos(x-\phi)$ form instead and mismatching the phase shift, leading to $\pi/12$ turning into a different fraction.

Forgetting the leading factor of $\dfrac12$ that comes from $R=2$ .

✓Final answer
$\dfrac{1}{2}\log\left(\tan\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)\right)+c$

ANSWER: C
AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$\displaystyle\int_{\sqrt{2}}^{2} \dfrac{x}{(x^3 - x^2 + x - 1)(x + 1)}\, dx =$
›Reveal solutionSolution
The cubic factors nicely, and the whole denominator collapses to $x^4-1$ , turning the integral into a standard $u=x^2$ substitution followed by a log-of-ratio evaluation. Answer: $\dfrac14\log\left(\dfrac95\right)$ .

Concept and Intuition

Cubic factorisation by grouping ( $x^3-x^2+x-1=x^2(x-1)+(x-1)=(x-1)(x^2+1)$ ) is the key simplification; once multiplied by the extra $(x+1)$ factor, the denominator becomes the recognisable difference-of-squares product $(x^2-1)(x^2+1)=x^4-1$ , and $\int x/(x^4-1)\,dx$ is a textbook $u=x^2$ substitution leading to a partial-fraction log form.

Step-by-Step Solution

Factor by grouping: $x^3-x^2+x-1=x^2(x-1)+1(x-1)=(x-1)(x^2+1)$ .

Denominator $=(x-1)(x^2+1)(x+1)=(x-1)(x+1)(x^2+1)=(x^2-1)(x^2+1)=x^4-1$ .

Integral $=\displaystyle\int_{\sqrt2}^{2}\frac{x}{x^4-1}dx$ . Let $u=x^2$ , $du=2x\,dx$ ; limits $x=\sqrt2\Rightarrow u=2$ , $x=2\Rightarrow u=4$ .

$=\dfrac12\displaystyle\int_2^4\frac{du}{u^2-1}=\dfrac12\cdot\dfrac12\log\left|\dfrac{u-1}{u+1}\right|\Big|_2^4=\dfrac14\left[\log\dfrac{3}{5}-\log\dfrac13\right]$ .

$=\dfrac14\log\left(\dfrac{3/5}{1/3}\right)=\dfrac14\log\left(\dfrac95\right)$ .

Common Mistakes

Failing to spot the cubic factorisation and instead attempting a messy direct partial-fraction decomposition of a quartic-degree denominator.

Sign/limit slip when substituting $u=x^2$ and forgetting to convert the limits.

✓Final answer
$\dfrac{1}{4}\log\left(\dfrac{9}{5}\right)$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$\displaystyle\int_0^{\pi/2} \dfrac{\sin x}{1+\cos x+\sin x}\,dx =$
›Reveal solutionSolution
This tests the classic $I=J$ symmetry trick combined with the Weierstrass substitution; the answer is (B).

Concept and Intuition

When an integral is symmetric under $x\to \pi/2-x$ (swapping sine and cosine roles) but the numerator only has one of them, pairing it with its "partner" integral (numerator = the other function) lets you add the two to get something tractable, using $I=J$ from the symmetry.

Step-by-Step Solution

Let $I=\displaystyle\int_0^{\pi/2}\frac{\sin x}{1+\sin x+\cos x}dx$ , $J=\displaystyle\int_0^{\pi/2}\frac{\cos x}{1+\sin x+\cos x}dx$ .

Substituting $x\to \pi/2-x$ in $I$ turns it into $J$ , so $I=J$ .

$I+J=\displaystyle\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sin x+\cos x}dx=\int_0^{\pi/2}\left[1-\frac{1}{1+\sin x+\cos x}\right]dx=\frac{\pi}{2}-K$ , where $K=\displaystyle\int_0^{\pi/2}\frac{dx}{1+\sin x+\cos x}$ .

Compute $K$ using the Weierstrass substitution $t=\tan(x/2)$ : $\sin x=\frac{2t}{1+t^2}$ , $\cos x=\frac{1-t^2}{1+t^2}$ , $dx=\frac{2dt}{1+t^2}$ . Then $1+\sin x+\cos x=\frac{2(1+t)}{1+t^2}$ , so the integrand simplifies to $\frac{dt}{1+t}$ .

Limits: $x:0\to\pi/2$ gives $t:0\to1$ . So $K=\int_0^1\frac{dt}{1+t}=\log 2$ .

Since $I=J$ , $2I=\frac{\pi}{2}-\ln2$ , so $I=\frac{\pi}{4}-\frac12\ln2$ .

Common Mistakes

Forgetting to check $I=J$ before using $2I=I+J$ .

Errors in the Weierstrass substitution algebra (sign of $1-t^2$ ).

✓Final answer
$\dfrac{\pi}{4}-\dfrac{1}{2}\log 2$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$\displaystyle\int \dfrac{x}{(1-x^2)\sqrt{2-x^2}}\,dx =$
›Reveal solutionSolution
Substitute $u=2-x^2$ , then $w=\sqrt u$ , reducing the integral to a standard $\int dw/(w^2-1)$ logarithm form. Answer matches option (C) after an absolute-value sign adjustment.

Concept and Intuition

The presence of $x\,dx$ alongside $2-x^2$ inside a square root strongly suggests the substitution $u=2-x^2$ (its differential is exactly $-2x\,dx$ ). The remaining $(1-x^2)$ factor is then re-expressed in terms of $u$ as well, turning the whole integral into a rational function of $u$ (or of $w=\sqrt u$ ), solvable by the standard $\int dw/(w^2-1)$ log formula.

Step-by-Step Solution

Let $u = 2-x^2$ . Then $du = -2x\,dx \Rightarrow x\,dx = -\dfrac{du}{2}$ .

Also $x^2 = 2-u$ , so $1-x^2 = 1-(2-u) = u-1$ .

The integral $\displaystyle\int\frac{x\,dx}{(1-x^2)\sqrt{2-x^2}}$ becomes $\displaystyle\int \frac{-du/2}{(u-1)\sqrt u} = -\frac12\int\frac{du}{(u-1)\sqrt u}$ .

Let $w=\sqrt u$ , so $u=w^2$ , $du=2w\,dw$ :

$-\frac12\int\frac{2w\,dw}{(w^2-1)w} = -\int\frac{dw}{w^2-1} = -\frac12\log\left|\frac{w-1}{w+1}\right|+c = \frac12\log\left|\frac{w+1}{w-1}\right|+c.$

Substitute back $w=\sqrt{2-x^2}$ :

$\frac12\log\left|\frac{\sqrt{2-x^2}+1}{\sqrt{2-x^2}-1}\right|+c.$

Multiplying numerator and denominator inside the absolute value by $-1$ gives $\dfrac{1+\sqrt{2-x^2}}{1-\sqrt{2-x^2}}$ with the same absolute value (sign flips in both places), matching option (C) exactly, including the $\tfrac12$ coefficient.

Common Mistakes

Dropping the $\tfrac12$ factor picked up from $x\,dx=-du/2$ — this rules out option (A), which has the right log expression but no $\tfrac12$ .

Not noticing that $\left|\dfrac{a+1}{a-1}\right| = \left|\dfrac{1+a}{1-a}\right|$ inside an absolute value, which is the key step that makes our derived form match option (C) rather than looking mismatched.

✓Final answer
$\dfrac{1}{2}\log\left|\dfrac{1+\sqrt{2-x^2}}{1-\sqrt{2-x^2}}\right| + c$

ANSWER: C
AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$\displaystyle\int_{1/2}^{1/\sqrt{2}} \dfrac{1}{(x+\sqrt{1-x^2})(1-x^2)}\,dx =$
›Reveal solutionSolution
The substitution $x=\sin\theta$ turns the awkward algebraic integrand into $\sec^2\theta/(1+\tan\theta)$ , a direct log integral. Answer: $\log(3-\sqrt3)$ .

Concept and Intuition

Whenever $\sqrt{1-x^2}$ appears together with $x$ in a rational expression, the trigonometric substitution $x=\sin\theta$ is the natural move — it turns $\sqrt{1-x^2}$ into $\cos\theta$ and $dx$ into $\cos\theta\,d\theta$ , often producing large cancellations.

Step-by-Step Solution

Let $x=\sin\theta$ , $dx=\cos\theta\,d\theta$ , $\sqrt{1-x^2}=\cos\theta$ , $1-x^2=\cos^2\theta$ .

Limits: $x=\tfrac12\Rightarrow\theta=\tfrac\pi6$ ; $x=\tfrac1{\sqrt2}\Rightarrow\theta=\tfrac\pi4$ .

The integral becomes

$\int \frac{\cos\theta\,d\theta}{(\sin\theta+\cos\theta)\cos^2\theta} = \int\frac{d\theta}{\cos\theta(\sin\theta+\cos\theta)} = \int\frac{d\theta}{\sin\theta\cos\theta+\cos^2\theta}.$

Divide numerator and denominator by $\cos^2\theta$ : $\displaystyle\int \frac{\sec^2\theta\,d\theta}{\tan\theta+1}$ .

Let $t=\tan\theta$ , $dt=\sec^2\theta\,d\theta$ : $\displaystyle\int\frac{dt}{t+1} = \log|t+1|$ .

Evaluate from $\theta=\pi/6$ ( $t=1/\sqrt3$ ) to $\theta=\pi/4$ ( $t=1$ ):

$\log(1+1) - \log\!\left(1+\frac{1}{\sqrt3}\right) = \log 2 - \log\!\left(\frac{\sqrt3+1}{\sqrt3}\right) = \ln2 + \log\sqrt3 - \log(\sqrt3+1) = \log\!\left(\frac{2\sqrt3}{\sqrt3+1}\right).$

Rationalize: $\dfrac{2\sqrt3}{\sqrt3+1}\cdot\dfrac{\sqrt3-1}{\sqrt3-1} = \dfrac{2\sqrt3(\sqrt3-1)}{2} = \sqrt3(\sqrt3-1) = 3-\sqrt3$ .

So the value is $\log(3-\sqrt3)$ .

Common Mistakes

Forgetting to rationalize the final expression, leaving it in a form that doesn't visibly match any option.

Losing track of the limits after the substitution $t=\tan\theta$ (mixing up $\theta$ -limits with $t$ -limits).

✓Final answer
$\log(3-\sqrt{3})$

ANSWER: D
AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$\displaystyle\int \dfrac{13\cos 2x - 9\sin 2x}{3\cos 2x - 4\sin 2x}dx =$
›Reveal solutionSolution
Splitting the numerator into a multiple of the denominator plus a multiple of the denominator's derivative reduces the integral to $3x - \tfrac12\log|3\cos2x-4\sin2x|+c$ .

Concept and Intuition

For $\int \dfrac{p\cos2x+q\sin2x}{a\cos2x+b\sin2x}\,dx$ , write the numerator as $A(a\cos2x+b\sin2x)+B\dfrac{d}{dx}(a\cos2x+b\sin2x)$ . Then the integral splits into $A\int dx$ (linear in $x$ ) plus $B\int \dfrac{D'}{D}dx = B\log|D|$ — a logarithm.

Step-by-Step Solution

Let $D=3\cos2x-4\sin2x$ , so $D'=-6\sin2x-8\cos2x$ .

Write $13\cos2x-9\sin2x = A\,D+B\,D'$ :

$\cos2x$ : $13=3A-8B$

$\sin2x$ : $-9=-4A-6B$

Solving: from the second, $4A+6B=9$ . Substituting $A=(13+8B)/3$ gives $50B=-25\Rightarrow B=-\tfrac12$ , then $A=3$ .

So the integrand $=\dfrac{3D-\tfrac12D'}{D}=3-\tfrac12\dfrac{D'}{D}$ .

Integrating: $\displaystyle\int\left(3-\frac12\frac{D'}{D}\right)dx = 3x-\frac12\log|D|+c = 3x-\frac12\log|3\cos2x-4\sin2x|+c$ .

Common Mistakes

Sign errors when solving the two linear equations for $A,B$ .

Forgetting the $\tfrac12$ factor that comes from $D'$ having coefficients twice the "natural" ones (since $D$ involves $2x$ ).

✓Final answer
$3x-\dfrac12\log|3\cos2x-4\sin2x|+c$

ANSWER: A
AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$k \in N$
›Reveal solutionSolution
The sum telescopes into a Riemann sum that converges to $\displaystyle\int_0^{k-1}\frac{dx}{1+x}=\log k$ .

Concept and Intuition

Sums of the form $\sum \dfrac1{n+j}$ over a range proportional to $n$ are classic Riemann-sum-in-disguise problems: factor out $\tfrac1n$ and recognise $\tfrac jn$ as the sample point of a definite integral.

Step-by-Step Solution

The sum runs over $j=1$ to $(k-1)n$ terms: $\displaystyle S=\sum_{j=1}^{(k-1)n}\frac1{n+j}=\frac1n\sum_{j=1}^{(k-1)n}\frac1{1+\frac jn}$ .

As $n\to\infty$ , this is the Riemann sum for $\displaystyle\int_0^{k-1}\frac{dx}{1+x}$ (sampling $x=j/n$ with step $1/n$ , from just above $0$ to $k-1$ ).

$\displaystyle\int_0^{k-1}\frac{dx}{1+x}=\log(1+x)\Big|_0^{k-1}=\log k-\log1=\log k$ .

Common Mistakes

Miscounting the upper limit of the Riemann sum as $k$ instead of $k-1$ (the sum stops at $kn$ , i.e. $j=(k-1)n$ ).

Forgetting to check with a simple case, e.g. $k=2$ : $\sum_{n+1}^{2n}\frac1i\to\log2$ , confirming the formula.

✓Final answer
$\log k$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$\displaystyle\int \dfrac{1}{\cos x}\left[\dfrac{1}{\sin x}-\dfrac{1}{\sin x+3\cos x}\right]dx=$
›Reveal solutionSolution
Simplifying the bracket first collapses the integrand to a single rational trig fraction, which becomes an easy partial-fraction integral in $\tan x$ .

Concept and Intuition

Combining the two fractions inside the bracket over a common denominator cancels the $\sin x$ term neatly, leaving a $\cos x$ in the numerator that cancels the outer $1/\cos x$ — a strong hint to simplify algebraically before attempting any substitution.

Step-by-Step Solution

$\dfrac{1}{\sin x}-\dfrac{1}{\sin x+3\cos x}=\dfrac{(\sin x+3\cos x)-\sin x}{\sin x(\sin x+3\cos x)}=\dfrac{3\cos x}{\sin x(\sin x+3\cos x)}$ .

Multiplying by the outer $\dfrac{1}{\cos x}$ : integrand $=\dfrac{3}{\sin x(\sin x+3\cos x)}$ .

Divide numerator and denominator by $\cos^2x$ : $=\dfrac{3\sec^2x}{\tan x(\tan x+3)}$ .

Let $u=\tan x$ , $du=\sec^2x\,dx$ : integral $=\displaystyle\int\frac{3}{u(u+3)}du$ .

Partial fractions: $\dfrac{3}{u(u+3)}=\dfrac1u-\dfrac{1}{u+3}$ , so the integral is $\log|u|-\log|u+3|+c=\log\left|\dfrac{u}{u+3}\right|+c$ .

Substituting back $u=\tan x$ : $\log\left|\dfrac{\tan x}{\tan x+3}\right|+c=\log\left|\dfrac{\sin x}{\sin x+3\cos x}\right|+c$ .

Common Mistakes

Attempting substitution before simplifying the bracket, leading to an unnecessarily messy integral.

Losing track of the factor from partial fractions and introducing a spurious $\frac13$ .

✓Final answer
$\log\left|\dfrac{\sin x}{\sin x+3\cos x}\right|+c$

ANSWER: D
AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
$\displaystyle\int \sec\left(x - \dfrac{\pi}{3}\right)\sec\left(x + \dfrac{\pi}{6}\right)dx =$
›Reveal solutionSolution
Because the two angles differ by exactly $\pi/2$ , $\sec A\sec B$ splits into $\tan B-\tan A$ , and integrating gives $\log\left|\frac{\cos(x-\pi/3)}{\cos(x+\pi/6)}\right|+c$ .

Concept and Intuition

Whenever you see $\sec(x-\alpha)\sec(x+\beta)$ where $(x+\beta)-(x-\alpha)$ is a constant, there's a standard trick: use $\sin\big((x+\beta)-(x-\alpha)\big)=\sin(x+\beta)\cos(x-\alpha)-\cos(x+\beta)\sin(x-\alpha)$ , divide by $\cos(x-\alpha)\cos(x+\beta)$ to turn the product-of-secants into a difference of tangents, which integrates easily to logs of cosines.

Step-by-Step Solution

Let $A=x-\dfrac\pi3$ , $B=x+\dfrac\pi6$ . Then $B-A = \dfrac\pi6+\dfrac\pi3=\dfrac\pi2$ , a constant.

Use the identity $\sin(B-A)=\sin B\cos A-\cos B\sin A$ . Dividing both sides by $\cos A\cos B$ :

$\frac{\sin(B-A)}{\cos A\cos B} = \tan B - \tan A.$

Since $\sin(B-A)=\sin(\pi/2)=1$ , we get $\dfrac{1}{\cos A\cos B}=\tan B-\tan A$ , i.e.

$\sec A\sec B = \tan B - \tan A = \tan\!\left(x+\frac\pi6\right) - \tan\!\left(x-\frac\pi3\right).$

Integrate term by term using $\displaystyle\int\tan u\,du = -\log|\cos u|+c$ (coefficient of $x$ in both angles is $1$ ):

$\int\sec A\sec B\,dx = -\log\left|\cos\!\left(x+\frac\pi6\right)\right| + \log\left|\cos\!\left(x-\frac\pi3\right)\right| + c.$

Combine the logs: $\log\left|\dfrac{\cos(x-\pi/3)}{\cos(x+\pi/6)}\right|+c$ .

Common Mistakes

Mixing up which angle goes in the numerator vs denominator inside the log (a sign/order slip flips to the reciprocal, which is a different, wrong-looking but structurally similar option).

Forgetting that the trick requires the angle-difference to come out to a constant, not a function of $x$ .

✓Final answer
$\log\left|\dfrac{\cos\left(x-\frac{\pi}{3}\right)}{\cos\left(x+\frac{\pi}{6}\right)}\right|+c$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
$\displaystyle\int \dfrac{x}{x\tan x + 1}\, dx = \log f(x) + k$
›Reveal solutionSolution
Recognising $x/(x\tan x+1)$ as $\frac{d}{dx}\log(x\sin x+\cos x)$ gives $f(x)=x\sin x+\cos x$ , so $f(\pi/4)=\dfrac{\pi+4}{4\sqrt2}$ .

Concept and Intuition

Many " $\int(\ldots)=\log f(x)+k$ " CET questions are really asking you to spot a function whose logarithmic derivative $f'/f$ equals the integrand. Instead of integrating from scratch, guess a plausible $f$ built from the pieces in the integrand ( $x$ , $\sin x$ , $\cos x$ ) and check its derivative.

Step-by-Step Solution

Guess $f(x)=x\sin x+\cos x$ (a natural combination given the $x\tan x$ in the denominator).

Differentiate: $f'(x)=\sin x+x\cos x-\sin x=x\cos x$ .

Form $\dfrac{f'(x)}{f(x)}=\dfrac{x\cos x}{x\sin x+\cos x}$ . Divide numerator and denominator by $\cos x$ :

$\frac{f'(x)}{f(x)}=\frac{x}{x\tan x+1},$
exactly the given integrand. So $\int\frac{x}{x\tan x+1}dx=\log|x\sin x+\cos x|+k$ , confirming $f(x)=x\sin x+\cos x$ .

4. Evaluate at $x=\pi/4$ : $\sin(\pi/4)=\cos(\pi/4)=\dfrac{\sqrt2}{2}$ , so
$f(\pi/4)=\frac\pi4\cdot\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\frac{\sqrt2}{2}\left(\frac\pi4+1\right)=\frac{\sqrt2}{2}\cdot\frac{\pi+4}{4}=\frac{\sqrt2(\pi+4)}{8}=\frac{\pi+4}{4\sqrt2}.$
Common Mistakes

Trying to integrate $x/(x\tan x+1)$ directly by substitution rather than recognising the logarithmic-derivative shortcut — much slower and error-prone.

Rationalising $\dfrac{\sqrt2(\pi+4)}{8}$ incorrectly instead of simplifying to $\dfrac{\pi+4}{4\sqrt2}$ .

✓Final answer
$\dfrac{\pi+4}{4\sqrt2}$

ANSWER: C

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