int {frac{{a{x^{ - 2}} + b{x^{ - 1}} + c}}{{{ | vedclass

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

Question

(D) To solve the integral $\int {\frac{{a{x^{ - 2}} + b{x^{ - 1}} + c}}{{{x^{ - 3}}}}} \,dx$,first simplify the integrand by dividing each term in the

Vedclass · Accepted Answer

$\frac{1}{2}a{x^2} + \frac{1}{3}b{x^3} + \frac{1}{4}c{x^4} + k$

Vedclass · Answer

(D) To solve the integral $\int {\frac{{a{x^{ - 2}} + b{x^{ - 1}} + c}}{{{x^{ - 3}}}}} \,dx$,first simplify the integrand by dividing each term in the numerator by ${x^{ - 3}}$: $\frac{{a{x^{ - 2}} + b{x^{ - 1}} + c}}{{{x^{ - 3}}}} = a{x^{ - 2 - ( - 3)}} + b{x^{ - 1 - ( - 3)}} + c{x^{ - ( - 3)}} = a{x^1} + b{x^2} + c{x^3}$. Now,integrate the expression with respect to $x$: $\int {(ax + b{x^2} + c{x^3})} \,dx = a\int x \,dx + b\int {{x^2}} \,dx + c\int {{x^3}} \,dx$. Using the power rule $\int {{x^n}} \,dx = \frac{{{x^{n + 1}}}}{{n + 1}} + k$,we get: $a(\frac{{{x^2}}}{2}) + b(\frac{{{x^3}}}{3}) + c(\frac{{{x^4}}}{4}) + k = \frac{1}{2}a{x^2} + \frac{1}{3}b{x^3} + \frac{1}{4}c{x^4} + k$. Thus,the correct option is $D$.

$\int {\frac{{a{x^{ - 2}} + b{x^{ - 1}} + c}}{{{x^{ - 3}}}}} \,dx = $

Similar Questions

$\int \frac{5(x^6+1)}{x^2+1} \, dx =$ (Where $C$ is a constant of integration.)

Find the following integral: $\int \frac{dx}{x^{2}-16}$

Find the following integral: $\int(2x - 3 \cos x + e^x) \, dx$

$\int \tan^4 x \, dx = $

$\int \frac{1}{x\sqrt{x^2 - 1}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module