Mathematics

# $\int {{{\sqrt {{x^2} - 4} }}dx}$

##### SOLUTION
$I=\int \sqrt{x^2-4}dx$
$I=\int \sqrt{x^2-2^2}dx$
$I=\dfrac{x}{2}\sqrt{x^2-4}-\dfrac{4}{2}\log|x+\sqrt{x^2-4}|+C$

$I=\dfrac{x}{2}\sqrt{x^2-4}-2\log|x+\sqrt{x^2-4}|+C$

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

#### Realted Questions

Q1 Subjective Hard
Solve
$\int { \dfrac { x+2 }{ \sqrt { { x }^{ 2 }+2x+3 } } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
$\int \sin^{-1} \ (\dfrac{2x +2}{\sqrt{4x^{2} +8x +13}}\ )dx$ =

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\displaystyle\int{\frac{(1+x^2)dx}{(1-x^2)\sqrt{1+x^2+x^4}}} =$
• A. $\displaystyle I=-\frac{1}{2\sqrt{3}}\log{\left|\frac{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}-\sqrt{5}}{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}+\sqrt{5}}\right|}+C$
• B. $\displaystyle I=-\frac{1}{4\sqrt{3}}\log{\left|\frac{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}-\sqrt{3}}{\sqrt{2x^2+\displaystyle\frac{1}{x^2}+1}+\sqrt{3}}\right|}+C$
• C. $\displaystyle I=-\frac{1}{2\sqrt{3}}\log{\left|\frac{\sqrt{2x^2+\displaystyle\frac{1}{2x^2}+1}-\sqrt{3}}{\sqrt{2x^2+\displaystyle\frac{1}{2x^2}+1}+\sqrt{3}}\right|}+C$
• D. $\displaystyle I=-\frac{1}{2\sqrt{3}}\log{\left|\frac{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}-\sqrt{3}}{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}+\sqrt{3}}\right|}+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate :
$\displaystyle \int tan^3 2x sec 2x dx$.

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 $\displaystyle f(x)>0 \forall x \in R$