Mathematics

# Evaluate : $I = \int {\dfrac{{2x}}{{{x^2} - 6x + 6}}dx}$

##### SOLUTION
$\int \dfrac { 2 x } { x ^ { 2 } - 6 x + 6 } d x$

put $x ^ { 2 } - 6 x + 6 = t$

$( 2 x - 6 ) d x = d t$
$\int\dfrac { 2 x } { x ^ { 2 } - 6 x + 6 } d x$
$= \int \dfrac { 2 x - 6 } { x ^ { 2 } - 6 x + 6 } d x + \int \dfrac { 6 } { x ^ { 2 } - 6 x + 6 } d x$

$= \int \dfrac { d t } { t } + \int \dfrac { 6 } { ( x - 3 ) ^ { 2 } - ( \sqrt { 3 } ) ^ { 2 } } d x$

$= \log \left( x ^ { 2 } - 6 x + 6 \right) + 6 \cdot \dfrac { 1 } { 2 \cdot \sqrt { 3 } } \log \left( \dfrac { x - 3 - \sqrt { 3 } } { x - 3 + \sqrt { 3 } } \right) + c$

$= \log \left( x ^ { 2 } - 6 x + 6 \right) + \sqrt { 3 } \log \left( \dfrac { x - 3 -\sqrt { 3 } } { x - 3 + \sqrt { 3 } } \right) + c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
The value of $\displaystyle \int _{ 0 }^{ \pi }{ \sin ^{ 50 }{ x } \cos ^{ 49 }{ x } dx }$ is
• A. $\dfrac{\pi}{4}$
• B. $\dfrac{\pi}{2}$
• C. $1$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value $\displaystyle \sqrt{2}\int\frac{\sin x dx}{\sin(x-\frac{\pi}{4})}$ is
• A. $x-\displaystyle \log|\sin(x-\frac{\pi}{4})|+C$
• B. $\displaystyle x- \log|\cos(x-\frac{\pi}{4})|+C$
• C. $\displaystyle x+ \log|\cos(x-\frac{\pi}{4})|+C$
• D. $\displaystyle x+ \log|\sin(x-\frac{\pi}{4})|+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\displaystyle f\left ( -x \right )+f\left ( x \right )= 0$ then $\displaystyle \int_{0}^{x}f\left ( t \right )$ dt is
• A. an odd function
• B. a periodic function
• C. none of these
• D. an even function

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int\frac{\sin x.\cos x}{1+\sin^{4}x}d_{X}=$
• A. $\tan (sin^{2}x) +c$
• B. $2\tan^{-1}(\sin^{2}x) +c$
• C. $\displaystyle \frac{1}{\sin x+\cos x}+c$
• D. $\displaystyle \frac{1}{2}\tan^{-1}(\sin^{2}x)+c$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$