Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

$\small \int_{1}^{2}\frac{dx}{\left ( x^{2}-2x+4 \right )^{\frac{3}{2}}} =\frac{k}{k+5}$ ,then k is equal to:

Option 1)
1

Option 2)
2

Option 3)
3

Option 4)
4

Answers (1)

best_answer

As learnt in concept

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

- wherein

Since $\int f(x)dx=\int f(t)dt=\int f(\theta )d\theta$ all variables must be converted into single variable , $\left ( t\, or\ \theta \right )$

$\int_{1}^{2}\frac{dx}{\left ( x^{2}-2x+4 \right )\frac{3}{2}}$

put t = x-1

$=>\int_{0}^{1}\frac{dt}{\left ( \left ( t+1 \right )^{2}-2\left ( t+1 \right )+4 \right )^{\frac{3}{2}}}$

$=> \int_{0}^{1} \frac{dt}{\left ( t^{2}+3 \right )^{\frac{3}{2}}}$

$t=\sqrt{3}\tan \Theta$

$dt=\sqrt{3}\sec ^{2}\Theta \:d\Theta$

$=>\int_{0}^{\tan ^{-1}\frac{1}{\sqrt{3}}} \frac{\sqrt{3}\sec^{2}\theta \:d\theta }{3\sqrt{3}\sec ^{3}\theta }$

$=>\frac{1}{3}\int_{0}^{\tan ^{-1}\frac{1}{\sqrt{3}}} \cos \theta \:d\theta$

$=\frac{1}{3}\left [ \sin \theta \right ]^{\tan^{-1}\frac{1}{\sqrt{3}} }_{0}$

$=\frac{1}{3}\times \frac{1}{2}$

$=\frac{1}{6}$

$=\frac{K}{K+5}$

$=> K=1$

Option 1)

1

Correct option

Option 2)

2

Incorrect option

Option 3)

3

Incorrect option

Option 4)

4

Incorrect option

Posted by

perimeter

View full answer

Crack JEE Main with "AI Coach"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

jee_ads

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2023 Session 2 City Intimation Slip Live Updates: NTA JEE Mains admit card at jeemain.nta.nic.in soon

anam-khan

JEE Main 2023 April Session: How e-libraries can help accelerate exam preparation

anam-khan

Fake videos on JEE Main 2023 session 2 admit card, city slip release date circulating, NTA warns

Latest Question