int x^2 e^{x^3} d x= . . . . . . . | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To evaluate the integral $I = \int x^2 e^{x^3} dx$,we use the method of substitution.Let $u = x^3$.Then,differentiating both sides with respect to

Vedclass · Accepted Answer

$\frac{1}{3} e^{x^3}+c$

Vedclass · Answer

(A) To evaluate the integral $I = \int x^2 e^{x^3} dx$,we use the method of substitution.Let $u = x^3$.Then,differentiating both sides with respect to $x$,we get $du = 3x^2 dx$,which implies $x^2 dx = \frac{1}{3} du$.Substituting these into the integral,we have:$I = \int e^u \cdot \frac{1}{3} du$$I = \frac{1}{3} \int e^u du$$I = \frac{1}{3} e^u + c$Substituting back $u = x^3$,we get:$I = \frac{1}{3} e^{x^3} + c$.Thus,the correct option is $A$.

$\int x^2 e^{x^3} d x=$ . . . . . . .

Similar Questions

$\int {\frac{{1 - {x^7}}}{{x(1 + {x^7})}}} \,dx$ equals

If $\int \frac{1}{x\left[(\log x)^2+4 \log x-1\right]} d x=A \log \left[\frac{\log x+B}{\log x+C}\right]+K$ where $K$ is the constant of integration,then

The integral $\int {\frac{{xdx}}{{2 - {x^2} + \sqrt {2 - {x^2}} }}} $ equals

$\int \frac{d x}{(x+a)^{\frac{9}{7}}(x-b)^{\frac{5}{7}}} = ?$

The value of $\int \frac{dx}{\sqrt{x}(x + 9)}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module