int frac{x , dx}{x^2 + 4x + 5} = | vedclass

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

Question

(D) Let $I = \int \frac{x \, dx}{x^2 + 4x + 5}$.We can rewrite the numerator as $x = \frac{1}{2}(2x + 4) - 2 = \frac{1}{2}(2x + 4) - 2$.So,$I = \int \

Vedclass · Accepted Answer

$\frac{1}{2}\log(x^2 + 4x + 5) - 2	an^{-1}(x + 2) + c$

Vedclass · Answer

(D) Let $I = \int \frac{x \, dx}{x^2 + 4x + 5}$.We can rewrite the numerator as $x = \frac{1}{2}(2x + 4) - 2 = \frac{1}{2}(2x + 4) - 2$.So,$I = \int \frac{\frac{1}{2}(2x + 4) - 2}{x^2 + 4x + 5} \, dx$.$I = \frac{1}{2} \int \frac{2x + 4}{x^2 + 4x + 5} \, dx - 2 \int \frac{dx}{(x+2)^2 + 1}$.For the first integral,let $t = x^2 + 4x + 5$,then $dt = (2x + 4) \, dx$.$I = \frac{1}{2} \int \frac{dt}{t} - 2 \int \frac{dx}{(x+2)^2 + 1^2}$.Using the standard integrals $\int \frac{1}{t} \, dt = \log|t|$ and $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$:$I = \frac{1}{2} \log|x^2 + 4x + 5| - 2 	an^{-1}(x + 2) + c$.

$\int \frac{x \, dx}{x^2 + 4x + 5} = $

Similar Questions

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

If $\int \sin 5x \cos 3x \; dx = - \frac{\cos 8x}{16} + A$,then $A = $

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

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

$\int \tan(3x - 5) \sec(3x - 5) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module