If int_1^2 frac{dx}{(x^2-2x+4)^{frac{3}{2}}} | vedclass

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

Question

(A) Let $I = \int_1^2 \frac{dx}{(x^2-2x+4)^{\frac{3}{2}}} = \int_1^2 \frac{dx}{((x-1)^2+3)^{\frac{3}{2}}}$.Substitute $x-1 = \sqrt{3} 	an 	heta$,so

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $I = \int_1^2 \frac{dx}{(x^2-2x+4)^{\frac{3}{2}}} = \int_1^2 \frac{dx}{((x-1)^2+3)^{\frac{3}{2}}}$.Substitute $x-1 = \sqrt{3} 	an 	heta$,so $dx = \sqrt{3} \sec^2 	heta \ d	heta$.When $x=1$,$	heta=0$. When $x=2$,$	an 	heta = \frac{1}{\sqrt{3}}$,so $	heta = \frac{\pi}{6}$.$I = \int_0^{\frac{\pi}{6}} \frac{\sqrt{3} \sec^2 	heta}{(3 \sec^2 	heta)^{\frac{3}{2}}} \ d	heta = \int_0^{\frac{\pi}{6}} \frac{\sqrt{3} \sec^2 	heta}{3\sqrt{3} \sec^3 	heta} \ d	heta = \frac{1}{3} \int_0^{\frac{\pi}{6}} \cos 	heta \ d	heta$.$I = \frac{1}{3} [\sin 	heta]_0^{\frac{\pi}{6}} = \frac{1}{3} (\sin \frac{\pi}{6} - \sin 0) = \frac{1}{3} 	imes \frac{1}{2} = \frac{1}{6}$.Given $\frac{k}{k+5} = \frac{1}{6}$,we have $6k = k+5$,which implies $5k = 5$,so $k = 1$.

If $\int_1^2 \frac{dx}{(x^2-2x+4)^{\frac{3}{2}}} = \frac{k}{k+5}$,then $k$ has the value

Similar Questions

Let $f_n = \int_0^{\frac{\pi}{2}} \left(\sum_{k=1}^n \sin^{k-1} x\right) \left(\sum_{k=1}^n (2k-1) \sin^{k-1} x\right) \cos x \, dx$,where $n \in N$. Then $f_{21} - f_{20}$ is equal to $...........$.

$\int_{0}^{5} \frac{d x}{x^{2}+2 x+10} = $

Let $\alpha \in (0,1)$ and $\beta = \log_{e}(1-\alpha)$. Let $P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots + \frac{x^n}{n}$ for $x \in (0,1)$. Then the integral $\int_{0}^{\alpha} \frac{t^{50}}{1-t} dt$ is equal to

$\int_{0}^{\frac{\pi}{4}} e^{\tan^2 \theta} \sin^2 \theta \tan \theta d\theta =$

$\int_0^a {x^4 \sqrt{a^2 - x^2}} \,dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module