Mathematics

# Evaluate the given integral.$\displaystyle \int { \cfrac { x+2 }{ \sqrt { { x }^{ 2 }-1 } } } dx$

##### SOLUTION
$I=\displaystyle\int{\dfrac{x+2}{\sqrt{{x}^{2}-1}}dx}$

$I=\displaystyle\int{\dfrac{x\,dx}{\sqrt{{x}^{2}-1}}}+2\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-1}}}$

$I=\dfrac{1}{2}\displaystyle\int{\dfrac{2x\,dx}{\sqrt{{x}^{2}-1}}}+2\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-1}}}$

Consider $\dfrac{1}{2}\displaystyle\int{\dfrac{2x\,dx}{\sqrt{{x}^{2}-1}}}$

Let $t={x}^{2}-1\Rightarrow\,dt=2x\,dx$

$\Rightarrow\,\dfrac{1}{2}\displaystyle\int{\dfrac{2x\,dx}{\sqrt{{x}^{2}-1}}}=\dfrac{1}{2}\displaystyle\int{\dfrac{dt}{\sqrt{t}}}$

$=\displaystyle\int{{t}^{-\frac{1}{2}}dt}$

$=\dfrac{{t}^{-\frac{1}{2}+1}}{-\dfrac{1}{2}+1}+c$

$=\dfrac{{t}^{\frac{1}{2}}}{\dfrac{1}{2}}+c$

$=2\sqrt{t}+c$

$=2\sqrt{{x}^{2}-1}+c$ where $t={x}^{2}-1$

Consider $2\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-1}}}$

We know that $\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-{a}^{2}}}}=\log{\left|x+\sqrt{{x}^{2}-{a}^{2}}\right|}+c$

$2\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-1}}}=2\log{\left|x+\sqrt{{x}^{2}-1}\right|}+c$

$\therefore\,I=\dfrac{1}{2}\displaystyle\int{\dfrac{2x\,dx}{\sqrt{{x}^{2}-1}}}+2\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-1}}}$

$=2\sqrt{{x}^{2}-1}+2\log{\left|x+\sqrt{{x}^{2}-1}\right|}+c$

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

#### Realted Questions

Q1 Single Correct Hard
The integral $\displaystyle \int x { \cos^{ -1 }\left(\displaystyle \frac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right) dx }$ is equal to :
(Note : $(x>0)$)
• A. $-x + (1+x^2) \cot^{-1} x+c$
• B. $-x-(1+x^{2})\tan^{-1}xc$
• C. $x-(1+x^2)\cot^{-1}x+c$
• D. $-x+(1+x^{2})\tan^{-1}x+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Evaluate: $\displaystyle\int{\frac {3x^2\, +\, 1}{(x^2\, -\, 1)^3}} dx$
• A. $\displaystyle C-\frac { x }{ { \left( { x }^{ 2 }+1 \right) }^{ 2 } }$
• B. $\displaystyle \frac { x }{ { \left( { x }^{ 2 }-1 \right) }^{ 2 } }$
• C. $\displaystyle \frac { x }{ { \left( { x }^{ 2 }+1 \right) }^{ 2 } }$
• D. $\displaystyle C-\frac { x }{ { \left( { x }^{ 2 }-1 \right) }^{ 2 } }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
If $\int_{a}^{b}|\sin x| d x=8$ and $\int_{0}^{a+b}|\cos x| d x=9$ , then find the value of $\int_{a}^{b} x \sin x d x$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
Value of $\int\limits_1^5 {\left( {\sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {\left( {x - 1} \right)} } } \right)dx}$
• A. $\frac{8}{3}$
• B. $\frac{{32}}{3}$
• C. $\frac{{34}}{3}$
• D. $\frac{{16}}{3}$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$