int frac{x^3-7 x+6}{x^2+3 x} ,d x= | vedclass

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

Question

(C) First, perform polynomial long division on the integrand $\frac{x^3-7x+6}{x^2+3x}$.Dividing $x^3-7x+6$ by $x^2+3x$ gives a quotient of $(x-3)$ and

Vedclass · Accepted Answer

$\frac{x^2}{2}-3 x+2 \log |x|+c$, where $c$ is a constant of integration.

Vedclass · Answer

(C) First, perform polynomial long division on the integrand $\frac{x^3-7x+6}{x^2+3x}$.Dividing $x^3-7x+6$ by $x^2+3x$ gives a quotient of $(x-3)$ and a remainder of $(2x+6)$.Thus, $\frac{x^3-7x+6}{x^2+3x} = x-3 + \frac{2x+6}{x^2+3x}$.Simplify the remainder term: $\frac{2x+6}{x^2+3x} = \frac{2(x+3)}{x(x+3)} = \frac{2}{x}$ for $x 
eq -3$.Now, integrate the expression: $\int (x-3+\frac{2}{x}) dx$.$= \int x dx - \int 3 dx + \int \frac{2}{x} dx$.$= \frac{x^2}{2} - 3x + 2 \log |x| + c$.

$\int \frac{x^3-7 x+6}{x^2+3 x} \,d x=$

Similar Questions

$\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx =$

$\int \frac{a^{\sqrt{x}}}{\sqrt{x}} dx = $

$\int(x-a)(x^{n-1}+x^{n-2}a+\ldots+a^{n-1})dx=$ (where $C$ is a constant of integration)

$\int (2\sin x + \frac{1}{x}) \, dx$ is equal to

$\int \sqrt{1+\cos x} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module