If int frac{x+5}{x^2+4x+5} dx = a log(x^2+4x+ | vedclass

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

Question

(A) Let $I = \int \frac{x+5}{x^2+4x+5} dx$. We express the numerator as $x+5 = \lambda(2x+4) + \mu$. Comparing the coefficients of $x$ and the constan

Vedclass · Accepted Answer

$(\frac{1}{2}, 3, 2)$

Vedclass · Answer

(A) Let $I = \int \frac{x+5}{x^2+4x+5} dx$. We express the numerator as $x+5 = \lambda(2x+4) + \mu$. Comparing the coefficients of $x$ and the constant terms on both sides: $1 = 2\lambda \implies \lambda = \frac{1}{2}$ $5 = 4\lambda + \mu \implies 5 = 4(\frac{1}{2}) + \mu \implies 5 = 2 + \mu \implies \mu = 3$. Thus,$I = \int \frac{\frac{1}{2}(2x+4) + 3}{x^2+4x+5} dx = \frac{1}{2} \int \frac{2x+4}{x^2+4x+5} dx + 3 \int \frac{1}{(x+2)^2 + 1^2} dx$. Using the formula $\int \frac{f'(x)}{f(x)} dx = \log|f(x)|$ and $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$: $I = \frac{1}{2} \log(x^2+4x+5) + 3 	an^{-1}(x+2) + C$. Comparing this with the given form $a \log(x^2+4x+5) + b 	an^{-1}(x+k) + C$,we get $a = \frac{1}{2}$,$b = 3$,and $k = 2$.

If $\int \frac{x+5}{x^2+4x+5} dx = a \log(x^2+4x+5) + b \tan^{-1}(x+k) + C$,then $(a, b, k)$ equals

Similar Questions

$\int \frac{1}{\cos x+\sqrt{3} \sin x} dx =$

$\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$ is equal to

$\int \frac{dx}{\cos x(1+\cos x)} = $

$\int \frac{\cos 2x - 1}{\cos 2x + 1} dx = $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module