Mathematics

# Evaluate:$\displaystyle \int \dfrac{1}{(2x - 7)} \sqrt{(x - 3)(x - 4)}\ dx$

##### SOLUTION
$I=\displaystyle \int { \dfrac { dx }{ \left( 2x-7 \right) } \sqrt { \left( x-3 \right) \left( x-4 \right) } }$

$=\displaystyle \int { \dfrac { \sqrt { \left( x-3 \right) \left( x-4 \right) } }{ \left( 2x-7 \right) } dx }$

$=\displaystyle \int { \dfrac { \sqrt { { x }^{ 2 }-3x-4x+12 } }{ 2x-7 } dx }$

$=\displaystyle \int { \dfrac { \sqrt { { x }^{ 2 }-7x+12 } }{ 2x-7 } dx }$

$=\displaystyle \int { \dfrac { \sqrt { { x }^{ 2 }-2.\dfrac { 7 }{ 2 } x+{ \left( \dfrac { 7 }{ 2 } \right) }^{ 2 }-{ \left( \dfrac { 7 }{ 2 } \right) }^{ 2 }+12 } }{ 2x-7 } dx }$

$=\displaystyle \int { \dfrac { \sqrt { { \left( x-\dfrac { 7 }{ 2 } \right) }^{ 2 }+12-\dfrac { 49 }{ 4 } } }{ 2x-7 } dx }$

$=\displaystyle \int { \dfrac { \sqrt { { \left( x-\dfrac { 7 }{ 2 } \right) }^{ 2 }-{ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 } } }{ 2\left( 1-\dfrac { 7 }{ 2 } \right) } dx }$

Let ${ \left( x-\dfrac { 7 }{ 2 } \right) }^{ 2 }-{ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }=t^2$
$\Rightarrow 2\left( x-\dfrac { 7 }{ 2 } \right) dx=2t\;dt$ $\Rightarrow dx=\dfrac { 2tdt }{ 2\left( x-\dfrac { 7 }{ 2 } \right) }$

$\Rightarrow I=\displaystyle \int { \dfrac { t.tdt }{ 2{ \left( x-\dfrac { 7 }{ 2 } \right) }^{ 2 } } }$

$=\displaystyle \dfrac { 1 }{ 4 } \int { \dfrac { t^{ 2 }dt }{ { \left( x-\dfrac { 7 }{ 2 } \right) }^{ 2 } } }$

$=\displaystyle \dfrac { 1 }{ 4 } \int { \dfrac { t^{ 2 }+\dfrac { 1 }{ 4 } -\dfrac { 1 }{ 4 } dt }{ { t }^{ 2 }+\dfrac { 1 }{ 4 } } }$

$=\displaystyle \dfrac { 1 }{ 4 } \int { 1dt-\dfrac { 1 }{ 4 } \int { \dfrac { 1 }{ t^{ 2 }+\dfrac { 1 }{ 4 } } dt } }$

$=\displaystyle \dfrac { 1 }{ 4 } t-\dfrac { 1 }{ 4\times \dfrac { 1 }{ 4 } } \tan ^{ -1 }{ \dfrac { t }{ \dfrac { 1 }{ 4 } } +c }$

$=\dfrac { 1 }{ 4 } t-\tan ^{ -1 }{ \dfrac { t }{ \dfrac { 1 }{ 4 } } +c }$
$=\dfrac { 1 }{ 4 } \sqrt { { x }^{ 2 }-7x+12 } -\tan ^{ -1 }{ 4\sqrt { \left( x-3 \right) \left( x-4 \right) } +c }$
Hence, the answer is $\dfrac { 1 }{ 4 } \sqrt { { x }^{ 2 }-7x+12 } -\tan ^{ -1 }{ 4\sqrt { \left( x-3 \right) \left( x-4 \right) } +c } .$

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Subjective Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Single Correct Hard
$\displaystyle \int\frac{\log(x+1)-\log x}{x(x+1)}dx=$
• A. $\displaystyle \log(x-1)\log x+\frac{1}{2}(\log(x-1))^{2}-\frac{1}{2}(\log x)^{2}+c$
• B. $\displaystyle \frac{1}{2}(\log(x+1))^{2}+\frac{1}{2}(\log x)^{2}-\log(x+1)\log x+c$
• C. $[(\displaystyle \log(1+\frac{1}{x})]^{2}+c$
• D. $-\displaystyle \frac{1}{2}[\log (1+\displaystyle \frac{1}{x})]^{2}+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value of $\displaystyle\int\limits_{-2}^{2}|x|\ dx$
• A. $2$
• B. $8$
• C. none of these
• D. $4$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
For x > 0, let f(x) = $\displaystyle \int_{1}^{x}{ \frac{log t}{1 + t}} dt$. Then f(x) + f$\displaystyle \left( \frac{1}{x} \right)$ is equal to:
• A. $\displaystyle \frac{1}{4} log x^2$
• B. log x
• C. $\displaystyle \frac{1}{4} (log x)^2$
• D. $\displaystyle \frac{1}{2} (log x)^2$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Show that $\displaystyle \int_{0}^{\dfrac {\pi}{2}}[2\log \sin x-\log (\sin 2x)]dx =\dfrac {\pi}{2}\log_{e}\left(\dfrac {1}{2}\right)$.

$\displaystyle \int { \sqrt { \tan ^{ }{ x } } \,\sec ^{ 4 }{ x } } dx$