Passage

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 $$\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$$
Mathematics

To evaluate $$\displaystyle\int{\frac{dx}{(x-1)\sqrt{-x^2+3x-2}}}$$ one of the most suitable substitution could be


ANSWER

$$\sqrt{-x^2+3x-2}=u(1-x)$$

$$\sqrt{-x^2+3x-2}=u(x-2)$$


SOLUTION
$$-{ x }^{ 2 }+3x-2=-{ x }^{ 2 }+2x+x-2=-x\left( x-2 \right) +1\left( x-2 \right) =\left( 1-x \right) \left( x-2 \right) $$ 
has real roots $$1$$ and $$2$$
So from case III: $$\sqrt { -{ x }^{ 2 }+3x-2 } =u(1-x) $$ or $$u(x-2)$$
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Multiple Correct Hard Published on 17th 09, 2020
Mathematics

$$\displaystyle\int{\frac{{(x+\sqrt{1+x^2})}^{15}}{\sqrt{1+x^2}}dx}$$ is equal to


ANSWER

$$\displaystyle\frac{{(x+\sqrt{1+x^2})}^{15}}{15}+C$$


SOLUTION
$$\displaystyle\int{\frac{{(x+\sqrt{1+x^2})}^{15}}{\sqrt{1+x^2}}dx}$$ 

$$\displaystyle=\int { \frac { { { (x+\sqrt { 1+x^{ 2 } } ) }(x+\sqrt { 1+x^{ 2 } } ) }^{ 14 } }{ \sqrt { 1+x^{ 2 } }  } dx } $$

Substitute $$(x+\sqrt { 1+x^{ 2 } } )=t$$

$$\displaystyle (1+\frac { 2x }{ 2\sqrt { 1+x^{ 2 } }  } )dx=dt$$

$$\displaystyle \frac { { { (x+\sqrt { 1+x^{ 2 } } ) } } }{ \sqrt { 1+x^{ 2 } }  } dx=dt$$

So, 
$$I=\displaystyle \int{{ t }^{ 14 }dt } $$

$$\displaystyle =\frac { { { { t }^{ 15 } } } }{ 15 } +C$$

$$I=\displaystyle \frac { { { { (x+\sqrt { 1+x^{ 2 } } ) }^{ 15 } } } }{ 15 } +C$$
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Single Correct Hard Published on 17th 09, 2020
Mathematics

If $$\displaystyle I=\int{\frac{dx}{x-\sqrt{9x^2+4x+6}}}$$ to evaluate $$I$$, one of the most proper substitution could be


ANSWER

$$\sqrt{9x^2+4x+6}=u\pm3x$$


SOLUTION
As $$9=m>0$$ then from case 1
$$\sqrt{9x^2+4x+6}=u\pm \sqrt{m}x=u\pm 3$$
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Single Correct Hard Published on 17th 09, 2020
Questions 203525
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Realted Questions

Q1 Single Correct Medium
Let $$f:R \to R,g:R \to R$$ be continuous functions. Then the value of indtegral 
$$\displaystyle \int\limits_{In\lambda }^{InI/\lambda } {\frac{{f\left( {\dfrac{{{x^2}}}{4}} \right)\left[ {f\left( x \right) - f\left( { - x} \right)} \right]}}{{g\left( {\dfrac{{{x^2}}}{4}} \right)\left[ {g\left( x \right) + g\left( { - x} \right)} \right]}}} dx$$


  • A. a non-zero constant
  • B. Zero
  • C. None of these
  • D. depend on $$\lambda $$

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Q2 Subjective Medium
Evaluate: $$\int (1-x)\sqrt {x}\ dx$$

Asked in: Mathematics - Integrals


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Q3 Passage Hard
Integral of form $$\int f\left ( x, \sqrt{ax^{2}+bx+c} \right )dx$$ can be evaluated with the help of Euler's substitution in the following manners:
(a) If $$a> 0$$, we put $$\sqrt{ax^{2}+bx+c}=t\pm x\sqrt{a}$$ or $$ax^{2}+bx+c=t^{2}\pm tx\sqrt{a}+ax^{2}$$ i.e. $$bx+c=t^{2}\pm 2tx\sqrt{a}$$
(b) If $$c> 0$$, we put $$\sqrt{ax^{2}+bx+c}=tx\pm \sqrt c$$ or $$ax+b=t^{2}x\pm 2tx\sqrt{c}$$
(c) If the trinomial $$ax^{2}+bx+c$$ has $$\alpha $$, $$\beta $$ as its real zero's
i.e. $$ax^{2}+bx+c=a\left ( x-\alpha  \right )\left ( x-\beta  \right )$$ then we put $$\sqrt{ax^{2}+bx+c}=t\left ( x-\alpha  \right )$$ or $$t\left ( x-\beta  \right )$$
On the basis of above information answer the following question:

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Q4 Single Correct Medium
If $$y=\sqrt [ 4 ]{ { x }^{ 4 }-1 } $$ then $$\int { \frac { dx }{ \sqrt [ 4 ]{ { x }^{ 4 }-1 }  }  } $$ is equal to
  • A. $$-\frac { 3 }{ 4 } [log|\frac { y+1 }{ y-1 } |-{ Tan }^{ -1 }(y)]+c$$
  • B. $$-\frac { 1 }{ 2 } [log|\frac { y+x-1 }{ y-x+1 } |+{ Tan }^{ -1 }(y)]+c$$
  • C. $$-\frac { 3 }{ 8 } [log|\frac { y-x }{ y+x } |-{ Tan }^{ -1 }(\frac { y }{ x } )]+c$$
  • D. $$\frac { 1 }{ 4 } log|\frac { y-x }{ y+x } |-\frac { 1 }{ 2 } { Tan }^{ -1 }(\frac { y }{ x } )+c$$

Asked in: Mathematics - Integrals


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Q5 Single Correct Hard
The value of the integral $$\displaystyle \int_{\frac {1}{3}}^1\frac {(x-x^3)^{\frac {1}{3}}}{x^4}dx$$
  • A. $$0$$
  • B. $$3$$
  • C. $$4$$
  • D. $$6$$

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