#### 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

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

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

##### 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

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

##### 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

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

##### 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
Subjects 9
Chapters 126
Enrolled Students 111

#### 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 $\lambda$

1 Verified Answer | Published on 17th 09, 2020

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

1 Verified Answer | Published on 17th 09, 2020

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:

1 Verified Answer | Published on 17th 09, 2020

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. $\frac { 1 }{ 4 } log|\frac { y-x }{ y+x } |-\frac { 1 }{ 2 } { Tan }^{ -1 }(\frac { y }{ x } )+c$

1 Verified Answer | Published on 17th 09, 2020

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. $6$