Find the following integral: int frac{x+2}{2 | vedclass

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

Question

(A) To evaluate $\int \frac{x+2}{2 x^{2}+6 x+5} d x$,we express the numerator as $A \frac{d}{d x}(2 x^{2}+6 x+5) + B$.$x+2 = A(4x+6) + B = 4Ax + (6A+B

Vedclass · Accepted Answer

$\frac{1}{4} \log |2 x^{2}+6 x+5|+\frac{1}{2} 	an ^{-1}(2 x+3)+ C$

Vedclass · Answer

(A) To evaluate $\int \frac{x+2}{2 x^{2}+6 x+5} d x$,we express the numerator as $A \frac{d}{d x}(2 x^{2}+6 x+5) + B$.$x+2 = A(4x+6) + B = 4Ax + (6A+B)$.Equating coefficients,$4A = 1 \implies A = \frac{1}{4}$ and $6A + B = 2 \implies 6(\frac{1}{4}) + B = 2 \implies B = 2 - \frac{3}{2} = \frac{1}{2}$.Thus,the integral becomes $\frac{1}{4} \int \frac{4x+6}{2x^2+6x+5} dx + \frac{1}{2} \int \frac{1}{2x^2+6x+5} dx$.Let $I_1 = \int \frac{4x+6}{2x^2+6x+5} dx = \log |2x^2+6x+5| + C_1$.For $I_2 = \int \frac{1}{2x^2+6x+5} dx = \frac{1}{2} \int \frac{1}{x^2+3x+\frac{5}{2}} dx = \frac{1}{2} \int \frac{1}{(x+\frac{3}{2})^2 + (\frac{1}{2})^2} dx$.Using $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a})$,we get $I_2 = \frac{1}{2} \cdot \frac{1}{1/2} 	an^{-1}(\frac{x+3/2}{1/2}) + C_2 = 	an^{-1}(2x+3) + C_2$.Combining these,the final result is $\frac{1}{4} \log |2 x^{2}+6 x+5|+\frac{1}{2} 	an ^{-1}(2 x+3)+ C$.

Find the following integral: $\int \frac{x+2}{2 x^{2}+6 x+5} d x$

Similar Questions

$\int \frac{\sin x \cdot \sec ^2 x-\tan x \cdot \sin x+\cos x}{(1-\cos 2 x)} d x=$

$\int \frac{\sin x+\sin ^3 x}{\cos 2 x} \,d x=A \cos x+B \log |f(x)|+c$ (where $c$ is a constant of integration). Then the values of $A, B$ and $f(x)$ are:

The integral $\int\left(\frac{x}{x \sin x+\cos x}\right)^{2} d x$ is equal to (where $C$ is a constant of integration)

$\int \frac{dx}{\sin^6 x + \cos^6 x} = $

$\int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module