If $\displaystyle\int \dfrac{3^{1/x}}{x^2}\,dx = k\left(3^{1/x}\right) + c$, then the value of $k$ is: (a) $-\dfrac{1}{\log 3}$ (b) $\log 3$ (c) $\dfrac{1}{\log 3}$ (d) $-\log 3$

If ∫ dfrac31/xx²dx = k(31/x) + c, then the value of k is: (a) -(1)/(log 3) (b) log 3 (c) (1)/(log 3) (d) -log

Question 124 of 129

\displaystyle\int \dfrac{3^{1/x}}{x^2}\,dx = k\left(3^{1/x}\right) + c

(a)

-\dfrac{1}{\log 3}

(b)

\log 3

(c)

\dfrac{1}{\log 3}

(d)

-\log 3

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Substituting $u=1/x$ converts the integral into a standard exponential integral.

Let $u=\dfrac{1}{x}$ , so $du=-\dfrac{1}{x^2}dx$ , i.e. $\dfrac{dx}{x^2}=-du$ .

Then $\displaystyle\int\dfrac{3^{1/x}}{x^2}dx=\int 3^u(-du)=-\int 3^u\,du=-\dfrac{3^u}{\log3}+c=-\dfrac{3^{1/x}}{\log3}+c$ . …

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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.If $\displaystyle\int \dfrac{3^{1/x}}{x^2}\,dx = k\left(3^{1/x}\right) + c$ , then the value of $k$ 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.If ∫31/xx2 dx=k(31/x)+c\displaystyle\int \dfrac{3^{1/x}}{x^2}\,dx = k\left(3^{1/x}\right) + c∫x231/x​dx=k(31/x)+c, then the value of kkk is:

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Q.If $\displaystyle\int \dfrac{3^{1/x}}{x^2}\,dx = k\left(3^{1/x}\right) + c$ , then the value of $k$ is:

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