$\int \dfrac{\log x}{x}\,dx,\ (x>0)$ is : (a) $\dfrac{2}{x^2}+c$ (b) $\dfrac{1}{2}(\log x)^2+c$ (c) $-\dfrac{2}{x^2}+c$ (d) $-\dfrac{1}{2}(\log x)^2+c$

∫ (log x)/(x)dx,(x>0) is : (a) (2)/(x²)+c (b) (1)/(2)(log x)²+c (c) -(2)/(x²)+c (d) -(1)/(2)(log x)²+c

Question 34 of 42

\int \dfrac{\log x}{x}\,dx,\ (x>0)

(a)

\dfrac{2}{x^2}+c

(b)

\dfrac{1}{2}(\log x)^2+c

(c)

-\dfrac{2}{x^2}+c

(d)

-\dfrac{1}{2}(\log x)^2+c

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Use the substitution $u=\log x$ ; the $\dfrac{1}{x}\,dx$ becomes $du$ , leaving a standard power integral. The answer is $\dfrac{1}{2}(\log x)^2+c$ , option (b).

Substitution. Let

$u=\log x \implies du=\frac{1}{x}\,dx.$

Rewrite the integral.

$\int \frac{\log x}{x}\,dx=\int u\,du.$

Integrate. …

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

CA Foundation 2025Set jan-20251 markMCQ
$\displaystyle\int (2x + 5)^7 \, dx$
›Reveal solutionSolution

$\int (2x+5)^7\,dx = \dfrac{(2x+5)^8}{2 \cdot 8} = \dfrac{(2x+5)^8}{16} + C$ .

Step 1 — Use the standard form for a linear substitution
$\int (ax+b)^n\, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$
Step 2 — Identify the constants
$a = 2, \quad b = 5, \quad n = 7$
Step 3 — Substitute
$\int (2x+5)^7\, dx = \frac{(2x+5)^{8}}{2 \times 8} = \frac{(2x+5)^8}{16} + C$
Why the other options are wrong: (B) divides only by 7 (raises to the wrong power AND drops the $\frac{1}{a}$ ); (C) and (D) wrongly expand $(2x+5)$ as $(2x^2+5x)$ , which is not the inner function. …
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CA Foundation 2024Set sep-20241 markMCQ
$\int \dfrac{1}{x(x^5 + 1)} \, dx$
›Reveal solutionSolution

Substitute u = x⁵ then use partial fractions: the integral = (1/5)·ln(x⁵/(x⁵+1)) + c.

Step 1 — Prepare for substitution

Multiply top and bottom by $x^4$ :
$\int \dfrac{1}{x(x^5+1)}\,dx = \int \dfrac{x^4}{x^5(x^5+1)}\,dx$
Step 2 — Substitute u = x⁵

With $u = x^5$ , $du = 5x^4\,dx$ , i.e. $x^4\,dx = \dfrac{du}{5}$ :
$= \dfrac{1}{5}\int \dfrac{du}{u(u+1)}$
Step 3 — Partial fractions and integrate
$\dfrac{1}{u(u+1)} = \dfrac{1}{u} - \dfrac{1}{u+1}$ $\dfrac{1}{5}\int\!\left(\dfrac{1}{u} - \dfrac{1}{u+1}\right)du = \dfrac{1}{5}\big[\ln u - \ln(u+1)\big] + c = \dfrac{1}{5}\ln\!\left(\dfrac{x^5}{x^5+1}\right) + c$
Why the other options are wrong: (A) omits the 1/5 factor; (C) uses 1/3 (wrong power); (D) inverts the ratio to (x⁵+1)/x⁵. Only (B) has both the 1/5 and the correct x⁵/(x⁵+1). …
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Q. $\int \dfrac{\log x}{x}\,dx,\ (x>0)$ is :

Concept understanding — Integration by Substitution

Step 1 — Use the standard form for a linear substitution

Step 2 — Identify the constants

Step 3 — Substitute

Step 1 — Prepare for substitution

Step 2 — Substitute u = x⁵

Step 3 — Partial fractions and integrate

Q.∫log⁡xx dx, (x>0)\int \dfrac{\log x}{x}\,dx,\ (x>0)∫xlogx​dx, (x>0) is :

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Q. $\int \dfrac{\log x}{x}\,dx,\ (x>0)$ is :

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