int (x^2 + 3x + 2) e^x dx = . . . . . . + | vedclass

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

Question

(C) To evaluate the integral $I = \int (x^2 + 3x + 2) e^x dx$,we use the integration by parts formula: $\int u dv = uv - \int v du$. Let $u = x^2 + 3x

Vedclass · Accepted Answer

$(x^2 + x + 1) e^x$

Vedclass · Answer

(C) To evaluate the integral $I = \int (x^2 + 3x + 2) e^x dx$,we use the integration by parts formula: $\int u dv = uv - \int v du$. Let $u = x^2 + 3x + 2$ and $dv = e^x dx$. Then $du = (2x + 3) dx$ and $v = e^x$. $I = (x^2 + 3x + 2) e^x - \int (2x + 3) e^x dx$. Now,apply integration by parts again for $\int (2x + 3) e^x dx$: Let $u_1 = 2x + 3$ and $dv_1 = e^x dx$. Then $du_1 = 2 dx$ and $v_1 = e^x$. $\int (2x + 3) e^x dx = (2x + 3) e^x - \int 2 e^x dx = (2x + 3) e^x - 2e^x = (2x + 1) e^x$. Substituting this back into the expression for $I$: $I = (x^2 + 3x + 2) e^x - (2x + 1) e^x + C$. $I = (x^2 + 3x + 2 - 2x - 1) e^x + C$. $I = (x^2 + x + 1) e^x + C$. Thus,the correct option is $C$.

$\int (x^2 + 3x + 2) e^x dx = $ . . . . . . $+ C$.

Similar Questions

The positive integer $n \leq 5$ for which $\int_0^1 e^x(x-1)^n dx = 16-6e$ is

$\int \frac{1}{x^2} \log(x^2 + a^2) \, dx = $

If $\int f(x) dx = \psi(x)$,then $\int x^5 f(x^3) dx = $

$\int \text{cosec}^{-1} \left( \sqrt{\frac{a + x}{x}} \right) dx =$

$\int \frac{\log x}{x^3} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module