Mathematics

# Evaluate  :        $\int {\frac{{x - 5}}{{\sqrt {{x^2} + 6x + 7} }}dx} .$

##### SOLUTION
$I=\quad \int { \cfrac { x-5 }{ \sqrt { { x }^{ 2 }+6x+7 } } } dx=\quad \int { \cfrac { x+3-8 }{ \sqrt { { x }^{ 2 }+6x+7 } } } dx=\int { \cfrac { x+3 }{ \sqrt { { x }^{ 2 }+6x+7 } } } dx-\int { \cfrac { 8 }{ \sqrt { { x }^{ 2 }+6x+7 } } } dx$
$A=\int { \cfrac { x+3 }{ \sqrt { { x }^{ 2 }+6x+7 } } } dx$
${ x }^{ 2 }+6x+7={ t }^{ 2 }$
$(2x+6)dx=2tdt\Rightarrow (x+3)dx=tdt$
$\quad A=\int { \cfrac { tdt }{ t } } \Rightarrow A=t+C\Rightarrow A=\sqrt { { x }^{ 2 }+6x+7 } +C$
$B=\int { \cfrac { 8 }{ \sqrt { { x }^{ 2 }+6x+7 } } } dx\quad \quad$
$x+3=p\Rightarrow dx=dp\quad$
$B=\int { \cfrac { 8dp }{ \sqrt { { p }^{ 2 }-{ (\sqrt { 2 } ) }^{ 2 } } } }$
$p=\sqrt { 2 } \sec { \theta } \Rightarrow dp=\sqrt { 2 } \sec { \theta } -\tan { \theta } d\theta$
$B=\int { \cfrac { 8\sqrt { 2 } \sec { \theta } -\tan { \theta } d\theta }{ \sqrt { 2\sec ^{ 2 }{ \theta } -2 } } } \quad =\int { \cfrac { 8\sec { \theta } -\tan { \theta } d\theta }{ \tan { \theta } } = } \int { 8 } \sec { \theta } d\theta =8\ln { \left| \sec { \theta } +\tan { \theta } \right| } =8\ln { \left| \cfrac { p }{ \sqrt { 2 } } +\sqrt { \cfrac { { p }^{ 2 } }{ 2 } -1 } \right| } =8\ln { \left| \cfrac { x+3 }{ \sqrt { 2 } } +\sqrt { \cfrac { { (x+3) }^{ 2 }-2 }{ \sqrt { 2 } } } \right| }$
$\therefore$ $I=\sqrt { { x }^{ 2 }+6x+7 } +8\ln { \left| \cfrac { x+3+\sqrt { { x }^{ 2 }+6x+7 } }{ \sqrt { 2 } } \right| } +C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

Q1 Subjective Medium
Evaluate :
$\int{ {e}^{x} \left[ \dfrac { 1 + \text{x log x}}{x} \right]} dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Solve $\int _{0}^{\pi} x\sec{x}\tan{x}dx$
• A. $-2\pi$
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• C. None of these
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1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium

$\displaystyle \int\frac{\sec x}{\sqrt{\sin(2x+\alpha)+\sin\alpha}}dx$ is equal to
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1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
The value of $\displaystyle \int_{-1}^{1}\left( \dfrac{x^2+sinx}{1+x^2}\right)dx$ is equal to
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1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
In calculating a number of integrals we had to use the method of integration by parts several times in succession.
The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts
$\displaystyle \int u\left ( x \right )v\left ( x \right )dx=u\left ( x \right )v_{1}-u'\left ( x \right )v_{2}\left ( x \right )+u''\left ( x \right )v_{3}\left ( x \right )+...+\left ( -1 \right )^{n-1}u^{n-1}\left ( x \right )V_{n}\left ( x \right ) \\ -\left ( -1 \right )^{n-1}\int u^{n}\left ( x \right )V_{n}\left ( x \right )dx$
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Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration by parts is especially useful when  calculating $\displaystyle \int P_{n}\left ( x \right )Q\left ( x \right )dx,$ where $\displaystyle P_{n}\left ( x \right )$ is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n+1 times.