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

Find: $\int \sin 2x \cos 3x \, dx$

If $\int (\sin 2x + \cos 2x) dx = \frac{1}{\sqrt{2}} \sin (2x - c) + a$,then the value of $a$ and $c$ is:

Find the following integral: $\int \sqrt{x}(3x^{2} + 2x + 3) dx$

If $\int {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}dx = \frac{1}{k}\left( {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}} \right)} + l$,then $k =$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module