3x^2 e^x^3 dx is equal to : (where C is an arbitrary constant) Option 1: 3x^3+C Option 2: 1/3e^x^3+C Option 3: e^x^3+C Option 4: 1/2e^x^2+C

Answer: e^x^3+C. Solution: Substitute t = x^3, so dt = 3x^2 dx. The integral becomes e^t dt = e^t + C. Back-substituting gives e^x^3 + C.

Calculus

Integrals

Easy

common

$\int 3x^2 e^{x^3}\,dx$ is equal to : (where $C$ is an arbitrary constant)

$3x^3+C$

$\dfrac{1}{3}e^{x^3}+C$

$e^{x^3}+C$

$\dfrac{1}{2}e^{x^2}+C$

✅ Correct Option: 3

Substitute $t = x^3$ , so $dt = 3x^2\,dx$ . The integral becomes $\int e^t\,dt = e^t + C$ .

Back-substituting gives $e^{x^3} + C$ .

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If $\int_0^1 \frac{1-x}{1+x} dx = A\log 2 + B$ , then the value of A and B are

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The value of which of the following integrals is zero?

(A) $\int_0^1 x dx$

(B) $\int_{-1}^1 x dx$

(D) $\int_0^1 \log\left(\frac{x}{1-x}\right) dx$

Choose the correct answer from the options given below:

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Match List-I with List-II

[.] denotes the greatest integer function.

Choose the correct answer from the options given below:

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If $g(x) = \int \frac{dx}{x^{1/2} + x^{1/6}}$ , then $g(1) - g(0)$ is :

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$\displaystyle\int \dfrac{dx}{(x^2+4)^{\frac{3}{2}}}$ is equal to ( where $c$ is an arbitrary constant)